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.
from math import sqrt, ceil, log
from heapq import heapify, heappush, heappop
from collections import defaultdict
from bisect import bisect_left, bisect_right

ioint = lambda: int(input())
ioarr = lambda: [int(i) for i in input().strip().split()]
prnys = lambda: print("YES")
prnno = lambda: print("NO")

MAX = 10**9

T = 1
# T = ioint()
while T:
    T -= 1

    n, q = ioarr()

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

    sub = [0] * (n + 1)
    for _ in range(q):
        l, r = ioarr()
        sub[l - 1] += 1
        sub[r] -= 1

    sm = [sub[0]] + [0] * n
    for i in range(1, n):
        sm[i] = sm[i - 1] + sub[i]
    sm.sort(reverse=True)

    res = 0
    for i in range(n):
        res += sm[i] * arr[i]
    print(res)
