CodeForces - 276C Little Girl and Maximum Sum （扫描线）

题意：

        有若干对区间和的查询，问如何重组数组使查询结果的和最大。N为区间长度，M为查询次数。

思路：

        统计出每一个位置出现的次数，统计的过程利用扫描线，然后当然出现最多的那个位置放最大的数，然后相乘累加就行

代码：

#include <bits/stdc++.h>

using namespace std;
const int MAXN=2e5+100;
long long a[MAXN];
long long lis[MAXN];
int main()
{
    ios::sync_with_stdio(false);
    long long n,m,st,en;
    while(cin>>n>>m){
        memset(lis,0,sizeof(lis));
        for(int i=0;i<n;i++)
            cin>>a[i];
        sort(a,a+n);
        while(m--){
            cin>>st>>en;
            lis[st-1]++;
            lis[en]--;
        }
        for(int i=1;i<n;i++)
            lis[i]+=lis[i-1];
        sort(lis,lis+n);
        long long ans=0;
        for(int i=0;i<n;i++)
            ans+=lis[i]*a[i];
        cout<<ans<<endl;
    }
}

The little girl loves the problems on array queries very much.

One day she came across a rather well-known problem: you've got an array of nelements (the elements of the array are indexed starting from 1); also, there are qqueries, each one is defined by a pair of integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n). You need to find for each query the sum of elements of the array with indexes from l_i to r_i, inclusive.

The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 2·10⁵) and q (1 ≤ q ≤ 2·10⁵) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers a_i (1 ≤ a_i ≤ 2·10⁵) — the array elements.

Each of the following q lines contains two space-separated integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the i-th query.

Output

In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Example

Input

3 3
5 3 2
1 2
2 3
1 3

Output

25

Input

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

Output

33