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.
totalNums, noOfQueries = map(int, input().split())
numsArray = list(map(int, input().split()))
numsArray.sort()

Queries = []
for idx in range(noOfQueries):
    Temp = list(map(int, input().split()))
    Queries.append(Temp)
Minimum, Maximum = 1, totalNums
# for id1 in range(noOfQueries):
#     Minimum = min(Minimum, Queries[id1][0])
#     Maximum = max(Maximum, Queries[id1][1])
Array = [0] * (Maximum + 1)
for id2 in range(noOfQueries):
    Array[Queries[id2][0] - 1] += 1
    Array[Queries[id2][1]] -= 1
for id3 in range(1, Maximum + 1):
    Array[id3] += Array[id3 - 1]
Array.pop()
Final = []
ty = [i for i in range(len(Array))]
temp = sorted(list(zip(Array, ty)))
final = [0] * (len(Array))
for k in range(len(Array)):
    final[temp[k][1]] = numsArray[k]
Sum = 0
for id5 in range(noOfQueries):
    Sum += sum(final[Queries[id5][0] - 1: Queries[id5][1]])
print(Sum)

# p = int(input("Enter no of elements:"))
# emptyList = []
# for idx in range(p):
#     emptyList.append(int(input()))
# for key in range(p):
#     emptyList[key] = emptyList[key] ** 2
# print(emptyList)
