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
input = sys.stdin.readline
n, q = map(int, input().split())
tree = [0] * (n + 1)
def read(x):
    sm = 0
    while x > 0:
        sm += tree[x]
        x -= x & -x
    return sm
def update(x, val):
    sm = 0
    while x <= n:
        tree[x] += val
        x += x & -x
def main():
    *arr, = sorted(map(int, input().split()))
    for _ in range(q):
        l, r = map(int, input().split())
        update(l, 1)
        update(r + 1, -1)
    pre = [0] * n
    for i in range(1, n + 1):
        pre[i - 1] = read(i)
    pre.sort()
    sm = 0
    for i in range(n):sm += arr[i] * pre[i]
    print(sm)
main()