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 = []
for _ in range (q):
    qr = list(map(int,input().split()))
    queries.append(qr)
nums.sort()

pre_sum = [0]*(length+1)

for query in queries:
    pre_sum[query[0]-1]+=1
    pre_sum[query[1]]-=1

prefix = [pre_sum[0]]
for i in range(1,length):
    count = pre_sum[i] + prefix[i-1]
    prefix.append(count)

index_priority = []
for i,pre in enumerate(prefix):
    indexs = (i,pre)
    index_priority.append(indexs)

index_priority.sort(key = lambda x : x[1] ,reverse=True)
ans = [0]*length
for idx in index_priority:
    ans[idx[0]] = nums.pop()

n_prefix = [ans[0]]
for idx in range(1,length):
    total = n_prefix[idx-1] + ans[idx]
    n_prefix.append(total)
result = 0
n_prefix.insert(0,0)

for qrs in queries:
    tot = n_prefix[qrs[1]] - n_prefix[qrs[0]-1]
    result+=tot
    
print(result)