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.Scanner;
import java.util.Comparator;
import java.util.Arrays;
import java.util.ArrayList;

public class Main {
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		long [] A = new long [N];
		int [][] BC = new int [M][2];
		for (int i = 0; i<N; i++) {
			A[i] = sc.nextLong();
		}
		Arrays.sort(A);
		for (int i = 0; i<M; i++) {
			BC[i][0] = sc.nextInt();
			BC[i][1] = sc.nextInt();
		}
		Arrays.sort(BC, new Comparator<int[]>() { //double[]でコンパレーターを実装する
			  public int compare(int[] a,int[] b) {
			    if(a[1] > b[1]) {
			      return 1;
			    }else if (a[1] == b[1]) {
			      return 0;
			    }else {
			      return -1;
			  }}});
		
		int tmp = M-1;
		int t = 0;
		Outer:
		while(BC[0][0] >=1 && t < N) {
			if(A[t] < BC[tmp][1]) {
				A[t] = BC[tmp][1];
				BC[tmp][0]--;
				t++;
				if(BC[tmp][0] == 0) {
					tmp--;
				}
			}
			else {
				break Outer;
			}
		}
		long sum = 0;
		for (int i =0; i< N;i++) {
			sum += A[i];
		}
		System.out.println(sum);
	}
}	

