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>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
  int N,M;
  cin >> N >> M;
  vector<pii> v(N+M);
  rep(i,N){
    int a;
    cin >> a;
    v[i]=pii(a,1);
  }
  rep(i,M){
    int b,c;
    cin >> b >> c;
    v[i+N]=pii(c,b);
  }
  sort(v.begin(),v.end(),greater<pii>());
  ll ans=0;
  int s=0;
  rep(i,M+N){
    s+=v[i].second;
    if(s<N) ans+=(ll)v[i].first*v[i].second;
    else{
      ans+=(ll)v[i].first*(v[i].second-(s-N));
      break;
    }
  }
  cout << ans << endl;
  return 0;
}