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.
length, q = list(map(int,input().split()))
nums = list(map(int,input().split()))

queries = []
pre_sum = [0]*(length+2)
for _ in range (q):
    qr = list(map(int,input().split()))
    pre_sum[qr[0]]+=1
    pre_sum[qr[1]+1]-=1
    queries.append(qr)
nums.sort(reverse=True)

for i in range(1,length+2):
    pre_sum[i] = pre_sum[i] + pre_sum[i-1]

index_priority = []
for j in range(1,length+1):
    indexs = (j,pre_sum[j])
    index_priority.append(indexs)
    
index_priority.sort(key = lambda x : x[1] ,reverse=True)
ans = [0]*(length+1)
for k,idx in enumerate(index_priority):
    ans[idx[0]] = nums[k]
    
for i in range(1,length+1):
    ans[i] = ans[i] + ans[i-1]
res = 0
for qrs in queries:
    tot = ans[qrs[1]] - ans[qrs[0]-1]
    res+=tot
print(res)

    
    
    
    
    
    

    
    