Problem: There are n balls. They are arranged in a row. Each ball has a color (for convenience an integer) and an integer value. The color of the i-th ball is ci and the value of the i-th ball is vi.

Squirrel Liss chooses some balls and makes a new sequence without changing the relative order of the balls. She wants to maximize the value of this sequence.

The value of the sequence is defined as the sum of following values for each ball (where a and b are given constants):

  * If the ball is not in the beginning of the sequence and the color of the ball is same as previous ball's color, add (the value of the ball)  ×  a. 
  * Otherwise, add (the value of the ball)  ×  b. 



You are given q queries. Each query contains two integers ai and bi. For each query find the maximal value of the sequence she can make when a = ai and b = bi.

Note that the new sequence can be empty, and the value of an empty sequence is defined as zero.

Input

The first line contains two integers n and q (1 ≤ n ≤ 105; 1 ≤ q ≤ 500). The second line contains n integers: v1, v2, ..., vn (|vi| ≤ 105). The third line contains n integers: c1, c2, ..., cn (1 ≤ ci ≤ n).

The following q lines contain the values of the constants a and b for queries. The i-th of these lines contains two integers ai and bi (|ai|, |bi| ≤ 105).

In each line integers are separated by single spaces.

Output

For each query, output a line containing an integer — the answer to the query. The i-th line contains the answer to the i-th query in the input order.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

6 3
1 -2 3 4 0 -1
1 2 1 2 1 1
5 1
-2 1
1 0


Output

20
9
4


Input

4 1
-3 6 -1 2
1 2 3 1
1 -1


Output

5

Note

In the first example, to achieve the maximal value:

  * In the first query, you should select 1st, 3rd, and 4th ball. 
  * In the second query, you should select 3rd, 4th, 5th and 6th ball. 
  * In the third query, you should select 2nd and 4th ball. 



Note that there may be other ways to achieve the maximal value.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Solution {
    class Input {
        int n;
        int q;
        long[] v;
        int[] c;
        long[] a;
        long[] b;
    }

    class Output {
        long[] ans;
    }

    Input loadData() throws Exception {
        Input inp = new Input();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        inp.n = Integer.parseInt(st.nextToken());
        inp.q = Integer.parseInt(st.nextToken());
        inp.v = new long[inp.n];
        inp.c = new int[inp.n];
        inp.a = new long[inp.q];
        inp.b = new long[inp.q];
        st = new StringTokenizer(br.readLine());
        for (int i=0; i<inp.n; i++) inp.v[i] = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        for (int i=0; i<inp.n; i++) inp.c[i] = Integer.parseInt(st.nextToken());
        for (int i=0; i<inp.q; i++) {
            st = new StringTokenizer(br.readLine());
            inp.a[i] = Integer.parseInt(st.nextToken());
            inp.b[i] = Integer.parseInt(st.nextToken());
        }
        br.close();
        return inp;
    }

    class TwoMaxStructure {
        long[] value;
        long max, max2;
        int maxPos, max2Pos;

        public TwoMaxStructure(int n) {
            value = new long[n+1];
        }

        void reset() {
            Arrays.fill(value, -1000000000000000000L);
            value[0] = 0;
            max = 0; maxPos = 0;
            max2 = value[1]; max2Pos = 1;
        }

        void update(int pos, long val) {
            if (value[pos] >= val) {
                return;
            }
            value[pos] = val;
            if (pos == maxPos) {
                max = val;
                return;
            }
            if (pos == max2Pos) {
                if (val >= max) {
                    max2Pos = maxPos; max2 = max;
                    maxPos = pos; max = val;
                } else {
                    max2 = val;
                }
                return;
            }
            if (val >= max) {
                max2Pos = maxPos; max2 = max;
                maxPos = pos; max = val;
            } else if (val >= max2) {
                max2Pos = pos; max2 = val;
            }
        }
    }

    Output solveProblem(Input inp) throws Exception {
        Output out = new Output();
        out.ans = new long[inp.q];
        TwoMaxStructure ds = new TwoMaxStructure(inp.n);
        for (int qq = 0; qq < inp.q; qq++) {
            long a = inp.a[qq], b = inp.b[qq];
            ds.reset();
            long ans = 0;
            for (int i=0; i < inp.n; i++) {
                long sameColor = ds.value[inp.c[i]] + inp.v[i] * a;
                long diffColor = (ds.maxPos != inp.c[i] ? ds.max : ds.max2) + inp.v[i] * b;
                long best = Math.max(sameColor, diffColor);
                ans = Math.max(ans, best);
                ds.update(inp.c[i], best);
            }
            out.ans[qq] = ans;
        }
        return out;
    }

    void printResult(Output out) throws Exception {
        for (long elm : out.ans) {
            System.out.println(elm);
        }
    }

    public void doMain() throws Exception {
        Input inp = loadData();
        Output out = solveProblem(inp);
        printResult(out);
    }

    public static void main(String[] args) throws Exception {
        new Solution().doMain();
    }
}