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.
def solve():
    n, q = map(int, input().split())
    tr = ct = 0
    v = [0] * (n + 2)
    temp = [0] * (n + 2)
    ans = list(map(int, input().split()))
    ans.sort(reverse=True)
    pq = []
    for i in range(q):
        x, y = map(int, input().split())
        pq.append((x, y))
        v[x] += 1
        v[y + 1] -= 1
    for i in range(1, n + 2):
        v[i] += v[i - 1]
    vp = [(v[i], i) for i in range(1, n + 1)]
    vp.sort(reverse=True)
    for i in range(len(vp)):
        temp[vp[i][1]] = ans[tr]
        tr += 1
    for i in range(1, n + 1):
        temp[i] += temp[i - 1]
    for i in range(len(pq)):
        ct += temp[pq[i][1]] - temp[pq[i][0] - 1]
    print(ct)

t = 1
for _ in range(t):
    solve()

		  		  	       		  			 					