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()))

count = [(0,j) for j in range(n)] 
queries = []
for i in range(q):
    l,r = map(int,input().split())
    queries.append([l,r])
    for k in range(l -1 ,r):
        a,b = count[k]
        a+=1
        count[k] = (a,b)
count.sort(key=lambda x: x[0])

arr.sort()
final = [0] * n
w = n - 1

while w >= 0:
    c,d = count[w]
    final[d] = arr[w]
    w-=1
tot = 0
pref =[0]
for i in range(len(final)):
    pref.append(pref[-1] + final[i])

for l,r in queries:
    tot+=(pref[r]-pref[l - 1])
print(tot)


