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

def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

res = 0
n,q = invr()
arr = inlt()
queries = []
for i in range(q):
    l,r = invr()
    queries.append((l,r))

temp = [0] * len(arr)
arr2 = [0] * len(arr)
for l,r in queries:
    temp[l-1] += 1

    if r < n:
        temp[r] -= 1

cur = 0
for i in range(n):
    cur += temp[i]
    arr2[i] = cur

arr.sort(reverse=True)
arr2.sort(reverse=True)

for i in range(n):res+= (arr[i] * arr2[i])

print(res)