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

def get_input():
    # Faster IO
    input_str = stdin.read().strip().split('\n')
    n, q = map(int, input_str[0].split())
    arr = list(map(int, input_str[1].split()))
    queries = [map(int, input_str[i].split()) for i in range(2, len(input_str))]

    return arr, queries

def get_max(arr, queries):
    n = len(arr)
    freq = [0] * (n + 2)

    for l, r in queries:
        freq[l] += 1
        freq[r + 1] -= 1

    for i in range(1, n + 1):
        freq[i] += freq[i - 1]

    freq.pop(0)
    freq.pop()
    freq.sort()
    arr.sort()
    # print(arr, freq)
    s = 0
    for i in range(n):
        s += arr[i] * freq[i]
    return s

print(get_max(*get_input()))
