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.
// NEEL
#include <cstdio>
#include <algorithm>

using namespace std;

int a[200000], c[200001] = {0};

int main()
{
    int n, q, l, r;
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; ++i)
    {
        scanf("%d", &a[i]);
    }
    while (q--)
    {
        scanf("%d%d", &l, &r);
        c[l-1] += 1;
        c[r] -= 1;
    }
    for (int i = 1; i < n; ++i)
    {
        c[i] += c[i-1];
    }

    sort(a, a + n);
    sort(c, c + n);
    long long answer = 0;
    for (int i = 0; i < n; ++i)
    {
        answer += static_cast<long long>(a[i]) * static_cast<long long>(c[i]);
    }
    printf("%I64d\n", answer);
    return 0;
}
	   	 	       		 		   	   	 	 	