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.
class FenwickTree:
    def __init__(self, x):
        """transform list into BIT"""
        self.bit = x
        for i in range(len(x)):
            j = i | (i + 1)
            if j < len(x):
                x[j] += x[i]

    def update(self, idx, x):
        """updates bit[idx] += x"""
        while idx < len(self.bit):
            self.bit[idx] += x
            idx |= idx + 1

    def query(self, end):
        """calc sum(bit[:end])"""
        x = 0
        while end:
            x += self.bit[end - 1]
            end &= end - 1
        return x

    def findkth(self, k):
        """Find largest idx such that sum(bit[:idx]) <= k"""
        idx = -1
        for d in reversed(range(len(self.bit).bit_length())):
            right_idx = idx + (1 << d)
            if right_idx < len(self.bit) and k >= self.bit[right_idx]:
                idx = right_idx
                k -= self.bit[idx]
        return idx + 1

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

tree = FenwickTree([0 for _ in range(n)])

for _ in range(q):
    l, r = list(map(int, input().split()))
    tree.update(l - 1, 1)
    tree.update(r, -1)

freq = []

for i in range(1, n + 1):
    freq.append(tree.query(i))

freq.sort()
a.sort()

t = 0

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

print(t)
