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.
#!/usr/bin/env python
# coding: utf-8

# In[14]:


from functools import cmp_to_key
def compare(p1:tuple,p2:tuple)->int:
    if p1[0]>p2[0]:
        return 1
    if p1[0]<p2[0]:
        return -1
    return 0


# In[18]:


n,q = map(int,input().split())
arr = [int(i) for i in input().split()]
count_arr = [0 for i in range(n)]
for i in range(q):
    li,ri = map(int,input().split())
    for i in range(li-1,ri):
        count_arr[i]+=1
        
from functools import cmp_to_key
def compare(p1:tuple,p2:tuple)->int:
    if p1[0]>p2[0]:
        return 1
    if p1[0]<p2[0]:
        return -1
    return 0
        
        
output = []
result = [0 for i in range(n)]
for i in range(n):
    temp_arr = []
    temp_arr.append(count_arr[i])
    temp_arr.append(i)
    output.append(temp_arr)
    
output.sort(key=cmp_to_key(compare))
arr.sort()
for i in range(n):
    result[output[i][1]] = arr[i]
    
summ = 0
for i in range(n):
    summ+=(count_arr[i]*result[i])
    
print(summ)
    


# In[ ]:


output = [[1,0],[3,1],[3,2],[1,3],[1,4]]
output.sort(key=cmp_to_key(compare))
print(output

