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;
#define ll long long

int main(){
  int N,M;
  cin>>N>>M;
  ll a=1,i=0,b;
  vector<pair<ll,int>> C(N+M);
  for(;i<N;i++){
    cin>>b;
    C[i]={b,a};
  }
  for(;i<N+M;i++){
    cin>>a>>b;
    C[i]={b,a};
  }
  ll ans=0;
  sort(C.begin(),C.end());
  i=M+N-1;
  for(;;i--){
    if(C[i].second<=N){
      ans+=C[i].first*C[i].second;
      N-=C[i].second;
    }else{
      ans+=N*C[i].first;
      break;
    }
  }
  cout<<ans<<endl;
}