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,q = map(int,input().split())
arr = list(map(int,input().split()))

c = [0]*n

sorted_arr = sorted(arr, reverse = True)
queries = []

for _ in range(q):
    a,b = map(int,input().split())
    queries.append((a-1,b-1))
    c[a-1]+=1
    if b<n:
        c[b]+=-1

for i in range(1,n):
    c[i]+=c[i-1]
sorted_c = []

for i,v in enumerate(c):
    sorted_c.append([v,i])
sorted_c = sorted(sorted_c,reverse = True)

temp = [0]*n
for idx,pack in enumerate(sorted_c):
    v,i = pack[0],pack[1]
    temp[i] = sorted_arr[idx]

for i in range(1,n):
    temp[i]+=temp[i-1]
total_sum = 0
for l,r in queries:
    if l==0:
        total_sum+=temp[r]
    else:
        total_sum+=temp[r]-temp[l-1]
print(total_sum)
