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.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

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();
        int[] B = new int[m], C = new int[m];
        for (int i = 0; i < m; i++) {
            B[i] = sc.nextInt();
            C[i] = sc.nextInt();
        }
        System.out.println(sln(n, m, A, B, C));
    }

    public static long sln(int n, int m, int[] A, int[] B, int[] C) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int a : A) pq.offer(a);
        int[][] D = new int[m][2];
        for (int i = 0; i < m; i++) {
            D[i][0] = B[i];
            D[i][1] = C[i];
        }
        Arrays.sort(D, new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                return b[1] - a[1];
            }
        });
        for (int i = 0; i < m; i++) {
            int k = D[i][0], v = D[i][1];
            for (int j = 0; j < k; j++) {
                int p = pq.peek();
                if (p < v) {
                    pq.poll();
                    pq.offer(v);
                } else break;
            }
        }
        long sum = 0;
        while (!pq.isEmpty()) {
            sum += (long) pq.poll();
        }
        return sum;
    }
}
