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
sys.setrecursionlimit(1000000000)
N = 200005
b = [0] * N
tree = [0] * N


def update(x, val):
    while x <= n:
        tree[x] += val
        x += (x & -x)


def su(x):
    s = 0
    while x > 0:
        s += tree[x]
        x -= (x & -x)
    return s


n, q = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
# print(niz)
a.sort()
for i in range(q):
    x, y = map(int, input().split())
    update(x, 1)
    update(y + 1, -1)
for i in range(1, n + 1):
    b[i - 1] = su(i)
b.sort()
for i in range(len(b)):
    if b[i] != 0:
        b = b[i::]
        break

ind = n - 1
while ind >= 0:
    ans += a[ind] * b[ind]
    ind -= 1
print(ans)
# print(new_niz)