CodeForces 377B Preparing for the Contest

Description

Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get c_i 'passes' in his subjects (regardless of the volume of his work).

Bugs, like students, are not the same: every bug is characterized by complexity a_j, and every student has the level of his abilities b_i. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: b_i ≥ a_j, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.

The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.

Input

The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 10⁵, 0 ≤ s ≤ 10⁹) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.

The next line contains m space-separated integers a₁, a₂, ..., a_m (1 ≤ a_i ≤ 10⁹) — the bugs' complexities.

The next line contains n space-separated integers b₁, b₂, ..., b_n (1 ≤ b_i ≤ 10⁹) — the levels of the students' abilities.

The next line contains n space-separated integers c₁, c₂, ..., c_n (0 ≤ c_i ≤ 10⁹) — the numbers of the passes the students want to get for their help.

Output

If the university can't correct all bugs print "NO".

Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.

Sample Input

Input

3 4 9
1 3 1 2
2 1 3
4 3 6

Output

YES
2 3 2 3

Input

3 4 10
2 3 1 2
2 1 3
4 3 6

Output

YES
1 3 1 3

Input

3 4 9
2 3 1 2
2 1 3
4 3 6

Output

YES
3 3 2 3

Input

3 4 5
1 3 1 2
2 1 3
5 3 6

Output

NO

Hint

Consider the first sample.

The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).

The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.

/**
     题意：m个bug, 第i个复杂度a[i],  n个学生,第i个能力b[i],花费c[i],
                 现在要解决掉所有bug,邀请一个学生i花费c[i],每个学生可以
                 解决掉复杂度小于能力值的bug,且每天最多解决掉一个，
                 问花费不超过s的情况下最少需要多少天能解决掉所有bug。
                 如果有解，输出“YES”并输出每个bug对应的解决者。
                 否则输出“NO”;

    分析：二分最少天数，贪心得出花费，检验是否少于s。
   */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N = 100010;
typedef long long LL;
struct Bug
{
    int id,a,ans;
    bool operator < (const Bug&cmp) const
    {
        return cmp.a<a;
    }
};
struct Stu
{
    int id,b,c;
    bool operator > (const Stu &cmp)const
    {
        return c > cmp.c;
    }
};
priority_queue<Stu , vector<Stu>, greater<Stu>  >q;
int cmp(Stu a, Stu b)
{
    return a.b < b.b;
}
Bug bugs[N];
Stu stu[N];
int str[N];
int main()
{
    //freopen("date.in","r",stdin);
    int i,j,k,m,n,s;
    while(scanf("%d %d %d",&n,&m,&s)==3)
    {
        for(i=1; i<=m; i++)
        {
            bugs[i].id = i;
            scanf("%d",&bugs[i].a);
        }
        for(i=1; i<=n; i++)
        {
            stu[i].id = i;
            scanf("%d",&stu[i].b);
        }
        for(i=1; i<=n; i++) scanf("%d",&stu[i].c);
        sort(bugs+1,bugs+1+m);
        sort(stu+1,stu+1+n,cmp);
        int l=1,r=m,ans=-1;
        while(l<=r)
        {
            int id = n;
            LL mony=0;
            while(!q.empty()) q.pop();
            bool ok = true;
            int mid = l+r>>1;
            for(i=1; i<=m; i+=mid)
            {
                while(id && stu[id].b >=bugs[i].a) q.push(stu[id]),id--;
                if(q.empty())
                {
                    ok = false;
                    break;
                }
                mony+=(LL)q.top().c;
                q.pop();
            }
            if(ok && mony<=(LL)s)
            {
                ans=mid, r = mid-1;
            }
            else l = mid+1;
        }
        if(ans==-1) puts("NO");
        else
        {
            puts("YES");
            int mid = ans;
            int id = n;
            while(!q.empty()) q.pop();
            for(i=1; i<=m; i+=mid)
            {
                while(id && stu[id].b >=bugs[i].a) q.push(stu[id]),id--;
                for(j=i; j<=m&&j<i+mid; j++)str[bugs[j].id] = q.top().id;
                q.pop();
            }
            for(i=1; i<m; i++) printf("%d ",str[i]);
            printf("%d\n",str[m]);
        }
    }
    return 0;
}