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 sys import stdin
inp = stdin.readline

n, qnr = map(int, inp().split())
arr = [int(x) for x in inp().split()]
q = []

for _ in range(qnr):
    a, b = map(int, inp().split())
    q.append((a-1, 1))
    q.append((b, 0))

q.sort()

active = 0
nr = 0
val = [0 for i in range(n)]

for c in q:
    while nr < c[0]:
        val[nr] = active
        nr += 1

    if c[1] == 1:
        active += 1
    else:
        active -= 1

val.sort()
arr.sort()

ans = 0
for i in range(n):
    ans += val[i]*arr[i]

print(ans)
