814/ A. An abandoned sentiment from past

题目链接：http://codeforces.com/problemset/problem/814/A

A. An abandoned sentiment from past

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.

To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.

Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length kequals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.

If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.

Input

The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.

The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.

Input guarantees that apart from 0, no integer occurs in a and b more than once in total.

Output

Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.

Examples

input

4 2
11 0 0 14
5 4

output

Yes

input

6 1
2 3 0 8 9 10
5

output

No

input

4 1
8 94 0 4
89

output

Yes

input

7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7

output

Yes

Note

In the first sample:

• Sequence a is 11, 0, 0, 14.
• Two of the elements are lost, and the candidates in b are 5 and 4.
• There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".

In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.

题意：给出a,b两个序列，b序列的个数就是a中0的个数（注意a中除0以外没有其他的数重，a,b中也没有重复的数），求把b中的数填充到a中（b中数只能用一次），使得a不递增

解析：特判k==1，其他直接输出Yes，晚上终测WA了，特判k==1出错

代码：

#include<bits/stdc++.h>
#define N 2009
using namespace std;

int a[N], b[N];

bool judge(int n, int k)
{
    if(k == 1)
    {
        if(a[1] == 0) a[1] = b[1];
        for(int i = 2; i <= n; i++)
        {
            if(a[i] == 0) a[i] = b[1];
            if(a[i] < a[i-1]) return true;
        }
        return false;
    }
    return true;
}

int main()
{
    int n, k;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = 1; i <= k; i++) scanf("%d", &b[i]);
    sort(b+1, b + k + 1);
    if(judge(n, k)) puts("Yes");
    else puts("No");
    return 0;
}