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 math
from collections import Counter, deque
from sys import stdout
import time
from math import factorial, log, gcd
import sys

def S():
    return sys.stdin.readline().split()


def I():
    return [int(i) for i in sys.stdin.readline().split()]


def II():
    return int(sys.stdin.readline())


class SegTree:
    def __init__(self, n):
        self.size = 1
        while self.size < n + 1:
            self.size *= 2
        self.tree = [0] * (2 * self.size - 1)

    def _set_item(self, i, v, x, lx, rx):
        if rx - lx == 1:
            self.tree[x] += v
            return
        m = (rx + lx) // 2
        if i < m:
            self._set_item(i, v, 2 * x + 1, lx, m)
        else:
            self._set_item(i, v, 2 * x + 2, m, rx)
        self.tree[x] = self.tree[2 * x + 1] + self.tree[2 * x + 2]

    def set_item(self, left, right):
        self._set_item(left - 1, 1, 0, 0, self.size)
        self._set_item(right, -1, 0, 0, self.size)

    def _get_item(self, left, right, x, lx, rx):
        if rx <= left or right <= lx:
            return 0
        if left <= lx and rx <= right:
            return self.tree[x]
        m = (rx + lx) // 2
        s1 = self._get_item(left, right, 2 * x + 1, lx, m)
        s2 = self._get_item(left, right, 2 * x + 2, m, rx)
        return s1 + s2

    def get_item(self, i):
        return self._get_item(0, i + 1, 0, 0, self.size)


def main():
    n, q = I()
    a = I()
    a.sort()
    tree = SegTree(n)
    for _ in range(q):
        left, right = I()
        tree.set_item(left, right)

    b = [0] * n
    for i in range(n):
        b[i] = tree.get_item(i)
    b.sort()
    s = 0
    for i in range(n):
        s += b[i] * a[i]
    print(s)


if __name__ == '__main__':
    # for _ in range(II()):
    #     main()
    main()