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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.AbstractCollection;
import java.util.PriorityQueue;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        IntegerCards solver = new IntegerCards();
        solver.solve(1, in, out);
        out.close();
    }

    static class IntegerCards {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();

            PriorityQueue<Long> pq = new PriorityQueue<>(n, Long::compare);
            // 降順
            PriorityQueue<Pair> conditions = new PriorityQueue<>(n, (o1, o2) -> Long.compare(o2.value, o1.value));
            for (int i = 0; i < n; i++) {
                pq.add(Long.parseLong(in.next()));
            }

            for (int i = 0; i < m; i++) {
                long b = Long.parseLong(in.next());
                long c = Long.parseLong(in.next());
                conditions.add(new Pair(c, b));
            }
            while (!conditions.isEmpty()) {
                Pair p = conditions.poll();
                for (int i = 0; i < p.time; i++) {
                    Long a = pq.poll();
                    if (p.value <= a) {
                        conditions.clear();
                        pq.add(a);
                        break;
                    }
                    pq.add(p.value);
                }
            }
            long result = 0;
            while (!pq.isEmpty()) {
                result += pq.poll();
            }
            out.print(result);
        }

        class Pair {
            public final Long value;
            public final Long time;

            public Pair(Long value, Long time) {
                this.value = value;
                this.time = time;
            }

        }

    }
}

