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 os
from collections import Counter


# Read from file.
if len(sys.argv) > 1:
    if os.path.exists(sys.argv[1]):
        file_name = sys.argv[1]
        with open(file_name, 'r') as f:
            input = iter(f.readlines()).__next__


# Read and store input.
n, q = map(int, input().split())
a = map(int, input().split())
ls, rs = list(), list()
for _ in range(q):
    l, r = map(int, input().split())
    ls.append(l)
    rs.append(r)

# Count how many times l and r were mentioned.
count_l = Counter(ls)
count_r = Counter(rs)

# Everytime an l is mentioned we increase with one, and everytime an r is mentioned we decrease with one for the next.
value = 0
idx_counts = dict()
for idx in range(1, n+1):
    value += count_l.get(idx, 0) - count_r.get(idx-1, 0)
    idx_counts[idx] = value

# Multiply the largest number with the most mentioned idx, etc.
ans = sum([ai*bi for ai, bi in zip(sorted(a, reverse=True), sorted(idx_counts.values(), reverse=True))])
print(ans)
