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, stdout


def next_line():
    return stdin.readline().rstrip()


def next_int():
    return int(next_line())


def out(text):
    stdout.write(str(text) + '\n')


def main():
    n, m = map(lambda x: int(x), next_line().split())
    arr = list(map(lambda x: int(x), next_line().split()))
    reqs = []
    for i in range(m):
        reqs.append(tuple(map(lambda x: int(x), next_line().split())))
    out(solve(arr, n, reqs))


def solve(arr, n, reqs):
    count_dif = [0] * (n + 1)
    for l, r in reqs:
        count_dif[l - 1] += 1
        count_dif[r] -= 1
    frequencies = [0] * n
    cur_count = 0
    for i in range(n):
        cur_count += count_dif[i]
        frequencies[i] = cur_count
    sorted_arr = sorted(arr, reverse=True)
    sorted_freq = sorted(frequencies, reverse=True)

    result = 0
    for i in range(n):
        result += sorted_arr[i] * sorted_freq[i]
    return result


main()
