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.
from array import *
nq=input()
x=nq.find(" ")
n=int(nq[0:x])
a = [0] * (n)
q=int(nq[x+1:])
tq = [""] * (q)
ch=input()
ch=ch+" "
for i in range(n):
    a[i]=int(ch[0:ch.find(" ")])
    ch=ch[ch.find(" ")+1:]
for i in range(q):
    tq[i]=input()
t= [0] * (n)
t2= [0] * (n)
for i in range(q):
    d=int(tq[i][0:tq[i].find(" ")])
    f=int(tq[i][tq[i].find(" ")+1:])
    for j in range(d-1,f):
        t[j]+=1
    t2[i]=i+1
a.sort(reverse=True)
for i in range(n):
        for j in range(0, n-i-1):
            if t[j] < t[j+1] :
                t[j], t[j+1] = t[j+1], t[j]
                t2[j], t2[j+1] = t2[j+1], t2[j]
m= [0] * (n)
for i in range(n):
    m[t2[i]-1]=a[i]
s=0
for i in range(q):
    d=int(tq[i][0:tq[i].find(" ")])
    f=int(tq[i][tq[i].find(" ")+1:])
    for j in range(d-1,f):
        s+=m[j]
print(s)