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.
t = [0 for _ in range(212345 * 4)]

def update(v, tl, tr, l, r):
    global t
    if l > r: return
    if l == tl and r == tr: t[v] += 1
    else:
        tm = (tl + tr) // 2
        update(v*2, tl, tm, l, min(r, tm))
        update(v*2+1, tm+1, tr, max(l, tm+1), r)

def get(v, tl, tr, pos):
    if tl == tr: return t[v]
    tm = (tl + tr) // 2
    if (pos <= tm): return t[v] + get(v*2, tl, tm, pos)
    else: return t[v] + get(v*2+1, tm+1, tr, pos)

n, q = list(map(int, input().split()))
nums = list(map(int, input().split()))
qs = []
for _ in range(q):
    a, b = input().split()
    qs.append([int(a), int(b)])
for p in qs: update(1, 1, n, p[0], p[1])
freqs = []
for i in range(1, n + 1): freqs.append(get(1, 1, n, i))
freqs.sort()
nums.sort()
sum = 0
for i in range(n): sum += freqs[i] * nums[i]
print(sum)


