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
import java.util.*;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int[] A = new int[N];
    for (int i = 0; i < N; i++) {
      A[i] = sc.nextInt();
    }
    Arrays.sort(A);
    
    TreeMap<Integer, Long> map = new TreeMap<>(Comparator.reverseOrder());
    for (int i = 0; i < M; i++) {
      int B = sc.nextInt();
      int C = sc.nextInt();
      map.put(C, map.getOrDefault(C,0L)+B);
    }
    
    long ans = 0;
    Iterator<Map.Entry<Integer, Long>> itr = map.entrySet().iterator();
    int i = N-1;
    long left = 0;
    while (i >= left && itr.hasNext()) {
      Map.Entry<Integer, Long> entry = itr.next();
      int val = entry.getKey();
      long count = entry.getValue();
      while (i >= left && val <= A[i]) {
        ans += A[i];
        i--;
      }
      if (i >= left && val > A[i]) {
        long len = Math.min((long) i-left+1, count);
        ans += val*len;
        left += len;
      }
    }
    while (i >= left) {
      ans += A[i];
      i--;
    }
    
    System.out.println(ans);
  }
}
