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.
n,m = map(int,input().split())
h = [*map(int,input().split())]
hsum = [0]*n

numCount = dict()
seq = []

for _ in [0]*m:
    a,b = map(int,input().split())
    seq.append([a,b])
    for i in range(a,b+1):
        if i in numCount:
            numCount[i] += 1
        else:
            numCount[i] = 1

numCount={k: v for k, v in sorted(numCount.items(), key=lambda item: item[1])}
u = reversed(numCount)
ans = [0]*n

for i in u:
    ans[i-1] = h.pop(h.index(max(h)))


hsum[0] = ans[0]

for i in range(1,n):
    hsum[i] = ans[i] + hsum[i-1]
hsum=[0]+hsum
res = 0
for i in seq:
    res += hsum[i[1]]-hsum[i[0]-1]
print(res)