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 $n$ elements (the elements of the array are indexed starting from 1); also, there are $q$ queries, each one is defined by a pair of integers $l_i$, $r_i$ $(1 \le l_i \le r_i \le n)$. 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.
The first line contains two space-separated integers $n$ ($1 \le n \le 2\cdot10^5$) and $q$ ($1 \le q \le 2\cdot10^5$) — the number of elements in the array and the number of queries, correspondingly.

The next line contains $n$ space-separated integers $a_i$ ($1 \le a_i \le 2\cdot10^5$) — the array elements.

Each of the following $q$ lines contains two space-separated integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the $i$-th query.
#include <bits/stdc++.h>

using namespace std;

int main(){
    int n, q;
    cin >> n >> q;
    vector<int> a(n);
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    sort(a.begin(), a.end());
    int b[n];
    for(int i = 0; i < n; i++){
        b[i] = 0;
    }
    for(int i = 0; i < q; i++){
        int l, r;
        cin >> l >> r;
        for(int j = l-1, j < r; j++){
            b[j]++;
        }
    }
    int k[q+1];
    for(int i = 0; i < q+1; i++){
        k[i] = 0;
    }
    long long ans = 0;
    for(int i = 0; i < n; i++){
        k[b[i]]++;
    }
    int index = q;
    for(int i = 0; i < n; i++){
        while (k[index] == 0){
            index -= 1;
        }
        ans += a[i] * index;
    }
    cout << ans << endl;
    return 0;
}
