codeforces276C

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 nelements (the elements of the array are indexed starting from 1); also, there are qqueries, each one is defined by a pair of integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ 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.

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 2·10⁵) and q (1 ≤ q ≤ 2·10⁵) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers a_i (1 ≤ a_i ≤ 2·10⁵) — the array elements.

Each of the following q lines contains two space-separated integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the i-th query.

Output

In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3 3
5 3 2
1 2
2 3
1 3

Output

25

Input

5 3
5 2 4 1 3
1 5
2 3
2 3

Output

33


正解是线段树，不过也有一种有趣的方法，计算每个点出现次数，将两个数组排序相乘即可

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<string.h>
 6 using namespace std;
 7 const int maxn = 21000;
 8 const int inf = 0x7f7f7f7f;
 9 const int mod = 1e9 + 7;
10 typedef long long ll;
11 typedef unsigned long long ull;
12 int n,q;
13 int a[maxn];
14 int b[maxn];
15 int main()
16 {
17     ios::sync_with_stdio(false);
18     cin>>n>>q;
19     for(int i=1;i<=n;i++)
20         cin>>a[i];
21     sort(a+1,a+1+n);
22     memset(b,0,sizeof(b));
23     int L,R;
24     for(int i=1;i<=n;i++){//神奇的求某个点的被加次数方法 
25         cin>>L>>R;
26         b[L]++;
27         b[R+1]--;
28     }
29     for(int i=1;i<=n;i++)
30         b[i]+=b[i-1]; 
31     sort(b+1,b+1+n);
32     ll ans=0;
33     for(int i=1;i<=n;i++)
34         ans+=a[i]*b[i];
35     cout<<ans<<endl;
36     return 0;
37  }