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=list(map(int,input().split()))
a = list(map(int,input().split()))
q = []
dist = [0] * (n+1)
for _ in range(m):
    l,r = list(map(int,input().split()))
    q.append([l,r])
    dist[l-1] += 1
    dist[r] -= 1
d = [(dist[0],0)]
for i in range(1,n):
    dist[i] += dist[i-1]
    d.append((dist[i],i))
d.sort(key=lambda item:item[0],reverse=True)
a.sort(reverse=True)
final = [0]*n

for i in range(n):
    final[d[i][1]] = a[i]

prefixSum = [0]

for i in range(n):
    prefixSum.append(prefixSum[-1] + final[i])
result = 0
for j in q:
    result += prefixSum[j[1]] - prefixSum[j[0]-1]
print(result)



