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 collections

if __name__ == '__main__':
    n, q = list(map(int, input().strip().split()))
    a = list(map(int, input().strip().split()))
    preSum = [0] * (n + 1)
    for i in range(q):
        l, r = list(map(int, input().strip().split()))
        preSum[l - 1] += 1; preSum[r] -= 1
    for i in range(1, n + 1):
        preSum[i] += preSum[i - 1]
    dic = collections.defaultdict(int)
    for i in range(n):
        dic[i] = preSum[i]
    dic = dict(sorted(dic.items(), key=lambda k: -k[1]))
    a.sort(reverse=True)
    pos, res = 0, 0
    for cur in dic:
        res += dic[cur] * a[pos]
        pos += 1
    print(res)