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.
import sys


def little_girl_and_maximum_sum(n: int, q: int):
    input = sys.stdin.readline
    arr = [int(x) for x in input().split()]
    d = [0 for _ in range(n)]
    for _ in range(q):
        left, right = map(int, input().split())
        d[left - 1] -= 1
        if right != n:
            d[right] += 1
    d[0] = -d[0]
    for i in range(1, n):
        d[i] = d[i - 1] - d[i]
    d.sort()
    arr.sort()
    res = 0
    for i in range(n):
        res += d[i] * arr[i]
    print(res)


def main():
    n, q = map(int, input().split())
    little_girl_and_maximum_sum(n, q)


if __name__ == '__main__':
    main()
