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 collections import defaultdict
n,q=map(int,input().split())
a=list(map(int,input().split()))
f=defaultdict(int)
inp=[]
for _ in range(q):
	b,c=map(int,input().split())
	inp.append([b-1,c-1])
	for m in range(b,c+1):
		if m-1 in f: f[m-1]+=1
		else: f[m-1]=1
f=list(sorted(f.items(),key=lambda x:x[1],reverse=True))
b=a.copy()
b.sort(key=lambda x:-x)
i=0
for j in range(len(f)):
	may=b[i]
	aux=a[f[j][0]]	
	a[f[j][0]]=a[a.index(may)]
	a[a.index(may)]=aux
	i+=1
pf=[0]*(n)
pf[0]=a[0]
for i in range(1,n):
	pf[i]=pf[i-1]+a[i]
sf=[0]*(n)
sf[n-1]=a[n-1]
for i in range(n-2,-1,-1):
	sf[i]=sf[i+1]+a[i]
suma=0
for k in range(q):
	i=inp[k][0]
	j=inp[k][1]
	#ahora hay que calcular la suma de una manera eficiente desde i hasta j
	if i-1>=0:
		if j+1<n:
			suma+=pf[n-1]-pf[i-1]-sf[j+1]
		else:
			suma+=pf[n-1]-pf[i-1]
	else:
		if j+1<n:
			suma+=pf[n-1]-sf[j+1]
		else:
			suma+=pf[n-1]
print(suma)
