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 collections import defaultdict

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

queries = []
ctr = defaultdict(int)
for i in range(q):
    start, end = list(map(int, input().split()))
    start -= 1
    end -= 1

    queries.append([start, end])
    ctr[start] += 1
    ctr[end + 1] -= 1

order = []
tmp = 0
for i in range(n):
    tmp += ctr[i]
    order.append(tmp)

arr = sorted([[index, i] for index, i in enumerate(arr)], key=lambda x: -order[x[0]])
arr = [i[1] for i in arr]

prefix = [0]
for i in arr:
    prefix.append(prefix[-1] + i)

res = 0
for start, end in queries:
    res += prefix[end+1] - prefix[start]
print(res)

