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 <iostream>
#include <algorithm>
#include <vector>

using namespace std;

const int constr = 1000006;

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());

  vector<int> b(n + 1, 0);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    b[l - 1] += 1;
    b[r] -= 1;
  }

  for (int i = 1; i < n; ++i) {
    b[i] += b[i - 1];
  }

  sort(b.begin(), b.end());

  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += a[i] * b[i];
  }

  cout << ans << endl;

  return 0;
}
