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 threading, sys

def main():
    n, q = map(int,input().split())
    nums = list(map(int, input().split()))
    spaces = [0]*n
    queries = []
    for _ in range(q):
        l,r = map(int, input().split())
        l-=1
        r-=1
        queries.append((l,r))
        spaces[l] += 1
        if r+1<n:
            spaces[r+1] -= 1

    for i in range(1, n):
        spaces[i] += spaces[i-1]
    
    spaces = sorted([(spaces[i], i) for i in range(n)], reverse=True)
    nums.sort(reverse=True)
    new = [0]*n

    for i in range(n):
        new[spaces[i][1]] = nums[i]
    
    p = [0]
    for num in new:
        p.append(p[-1]+num)
    
    tot = 0
    for l,r in queries:
        tot += p[r+1] - p[l]
    
    print(tot)


if __name__ == '__main__':
    sys.setrecursionlimit(1 << 30)
    threading.stack_size(1 << 27)

    main_thread = threading.Thread(target=main)
    main_thread.start()
    main_thread.join()