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.
def find_sum(bit,idx):
    num=0
    while idx>0:
        num+=bit[idx-1]
        idx-=idx&-idx
    return num
def update_bit(b,l,r,val):
    
    idx=l
    while idx<=r:
        b[idx-1]+=val
        idx+=idx&-idx
    
a = list(map(int,input().strip().split()))
n,q=a
a = list(map(int,input().strip().split()))
a.sort()
b=[0 for _ in range(n)]

for i in range(q):
    c = list(map(int,input().strip().split()))
    l,r=c
    update_bit(b,l,n,1)
    update_bit(b,r+1,n,-1)
# c=[b[0]]
c=[]
for i in range(1,n+1):
    # c.append(find_sum(b,i+1)-find_sum(b,i))
    c.append(find_sum(b,i))
c.sort()
num=0
for i in range(n):
    num+=a[i]*c[i]
print(num)