276_C. Little Girl and Maximum Sum
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 ≤ l_i ≤ r_i ≤ 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.

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 2⋅10^5) and q (1 ≤ q ≤ 2⋅10^5) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers a_i (1 ≤ a_i ≤ 2⋅10^5) — the array elements.

Each of the following q lines contains two space-separated integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the i-th query.

Output

In a single line print, a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3 3
5 3 2
1 2
2 3
1 3


Output

25


Input

5 3
5 2 4 1 3
1 5
2 3
2 3


Output

33
```python
n,m = map(int,input().split())
a = list(map(int,input().split()))
l = [0]*(n+2)
for i in range(m):
	x,y=map(int,input().split())
	l[x] += 1
	l[y+1] -= 1
 
for i in range(2, n+2):
	l[i] += l[i-1]
 
l.sort(reverse=True)
a.sort(reverse=True)
# print(l, a)
ans=0
for i in range(n):
	ans += l[i]*a[i]
 
print(ans)

```