You have N cards. On the i-th card, an integer A_i is written.

For each j = 1, 2, ..., M in this order, you will perform the following operation once:

Operation: Choose at most B_j cards (possibly zero). Replace the integer written on each chosen card with C_j.

Find the maximum possible sum of the integers written on the N cards after the M operations.

Constraints

* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i, C_i \leq 10^9
* 1 \leq B_i \leq N

Input

Input is given from Standard Input in the following format:


N M
A_1 A_2 ... A_N
B_1 C_1
B_2 C_2
\vdots
B_M C_M


Output

Print the maximum possible sum of the integers written on the N cards after the M operations.

Examples

Input

3 2
5 1 4
2 3
1 5


Output

14


Input

10 3
1 8 5 7 100 4 52 33 13 5
3 10
4 30
1 4


Output

338


Input

3 2
100 100 100
3 99
3 99


Output

300


Input

11 3
1 1 1 1 1 1 1 1 1 1 1
3 1000000000
4 1000000000
3 1000000000


Output

10000000001
#include <bits/stdc++.h>
using namespace std;

int main(){
  int N,M; cin >> N >> M;
  vector<pair<int,int>> A;
  for(int i=0; i<N; i++){
    int a; cin >> a;
    A.push_back(make_pair(a,1));
  }
  for(int i=0; i<M; i++){
    int b,c; cin >> b >> c;
    A.push_back(make_pair(c,b));
  }
  sort(A.begin(),A.end(),greater<pair<int,int>>());
  
  int i=0;
  long long sum=0;
  while(N){
    long long num=A[i].first,times=A[i].second;
    if(N>times){
      sum+=num*times;
      N-=times;
    }else{
      sum+=num*N;
      N=0;
    }
    i++;
  }
  cout << sum << endl;
}