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 
import math
from functools import cmp_to_key
from collections import OrderedDict
 
 
sys.setrecursionlimit(10**9)
 
 
 
 
 
def solve():


    n,q = list(map(int,input().split()))

    arr = [int(x) for x in input().split()]


    count = [0 for i in range(n+1)]

    for i in range(q):

        l,r = list(map(int,input().split()))

        count[l-1] += 1 
        count[r] -= 1



    for i in range(1,n+1):

        count[i] += count[i-1]

    count.pop()
    count.sort()
    arr.sort()

    ans = 0 

    i = n - 1

    while i >= 0:

        ans += arr[i]*count[i]
        i -= 1 

    print(ans)




 
 
 
 
  
 
if __name__ == '__main__':
 
 
    solve()