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.Arrays;
import java.util.Scanner;

public class Main {
	Scanner sc = new Scanner(System.in);
	
	public static void main(String[] args) {
		Main m = new Main();
		m.run();
	}
	
	void run() {
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		int a[] = new int[n];
		int cb[][] = new int[m][2];
		

		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		
		for(int i =0; i < m ;i++) {
			int b = sc.nextInt();
			int c = sc.nextInt();
			
			cb[i][0] = c;
			cb[i][1] = b;
		}
		
		Arrays.sort(a);
		Arrays.sort(cb, (arr1,arr2) -> - Integer.compare(arr1[0], arr2[0]));
		
		int b2[] = new int[n];
		
		int j = 0;
		for(int i = 0 ;i < m ;i++) {
			for(int k = 0; k < cb[i][1]; k++) {
				if(j >= n) {
					break;
				}
				b2[j] = cb[i][0];
				j++;
			}
			if(j>=n) {
				break;
			}
		}
		
		int c[] = new int[n+j];
		
		for(int x = 0; x < n ;x++) {
			c[x] = a[x];
		}
		for(int x = 0; x < j ; x++) {
			c[x+n] = b2[x];
		}
		
		Arrays.sort(c);
		
		long sum =0;
		
		for(int x = j ; x < n+j;x++) {
			sum += c[x];
		}
		System.out.println(sum);
	}
}
