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/python3

#read all input
n, q = map(int, input().split())
a = list(map(int, input().split()))
queries = []
for i in range(0, q):
    queries.append(list(map(int, input().split())))

#compute frequencies of index usage
class IndexFrequency:
  def __init__(self, index, frequency):
    self.index = index
    self.frequency = frequency

freqs = []
for i in range(0, n):
   freqs.append(IndexFrequency(i, 0))

for i in range(0, q):
    l = queries[i][0]
    r = queries[i][1]
    for j in range(l, r + 1):
        freqs[j - 1].frequency += 1

#sort freqs from highest
a.sort(reverse=True)
freqs.sort(key=lambda x:x.frequency, reverse=True)

#create final array
f = [0] * n
for i in range(0, n):
   f[freqs[i].index] = a[i]

#create sums array for quick counting
sum = 0
s = [0] * n
for i in range(0, n):
   sum += f[i]
   s[i] = sum
    
#count final max sum
sum = 0
for i in range(0, q):
    li = queries[i][0] - 1
    ri = queries[i][1] - 1
    sum += s[ri] - (s[li-1] if li > 0 else 0)

#print result
print(sum)