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();
		Queue<Long> pq = new PriorityQueue<Long>(N);
		for (int i = 0; i < N; i++) {
			pq.add(sc.nextLong());
		}
		Map<Long, Integer> tm = new TreeMap<Long, Integer>(Collections.reverseOrder());
		for (int i = 0; i < M; i++) {
			int b = sc.nextInt();
			long c = sc.nextLong();
			// treeMapに値を入れる
			if (tm.containsKey(c)) {
				int prev = tm.get(c);
				tm.put(c, prev + b);
			} else {
				tm.put(c, b);
			}
		}
		sc.close();

		tm.forEach((k, v) -> {
			int cnt = 0;
			while (true) {
				if (k > pq.peek()) {
					pq.poll();
					pq.add(k);

					cnt++;
					if (cnt == v) {
						return;
					}
				} else {
					break;
				}
			}
		});

		long ans = 0;
		for (int i = 0; i < N; i++) {
			ans += pq.poll();
		}
		System.out.println(ans);
	}
}
