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 static java.lang.Math.*;
import static java.util.Arrays.*;
import java.util.*;
import java.io.*;

public class Main {
	public static void main(String[] args) throws Exception {
		Class<?> here = new Object(){}.getClass().getEnclosingClass();
		try {
			String packageName = here.getPackage().getName();
			packageName = "src/" + packageName.replaceAll("\\.", "/") + "/";
			System.setIn(new FileInputStream(packageName + "input.txt"));
//			System.setOut(new PrintStream(new FileOutputStream(packageName + "output.txt")));
		} catch (FileNotFoundException e) {
		} catch (NullPointerException e) {
		}

		Object o = Class.forName(here.getName()).newInstance();
		o.getClass().getMethod("run").invoke(o);
	}

	static void tr(Object... os) {
		System.err.println(deepToString(os));
	}


	MyScanner sc = null;
	PrintWriter out = null;
	public void run() throws Exception {
		sc = new MyScanner(System.in);
		out = new PrintWriter(System.out);
		for (;sc.hasNext();) {
			solve();
			out.flush();
		}
		out.close();
	}

	void solve() {
		int n = sc.nextInt();
		int q = sc.nextInt();

		int v[] = sc.nextIntArray(n);
		int c[] = sc.nextIntArray(n);
		for (int i = 0; i < n; i++) c[i]--;

		for (; q --> 0; ) {
			int a = sc.nextInt();
			int b = sc.nextInt();

			long res = calc(a, b, v, c);
			out.println(res);

		}
	}

	long calc(long a, long b, int[] v, int []c) {
		int n = v.length;
		long [] maxColor = new long[n]; // color i で終わる最高点

		long inf = 1L << 58;
		fill(maxColor, -inf);

		long [][] other = new long [2][2]; // { color, score } のベスト２

		long start = v[0] * b;
		maxColor[c[0]] = start;

		other[0][0] = c[0];
		other[0][1] = start;

		other[1][0] = -inf;

		for (int i = 1; i < n; i++) {
//			tr(other);
			long here = v[i] * b;
//			here = max(here, v[i] * b);
			here = max(here, maxColor[c[i]] + v[i] * a);

			if (other[0][0] != c[i]) {
				here = max(here, other[0][1] + v[i] * b);
			}
			if (other[1][0] != c[i] && other[1][0] >= 0) {
				here = max(here, other[1][1] + v[i] * b);
			}

//			tr(i, v[i], c[i], here);

			// update
			maxColor[c[i]] = max(maxColor[c[i]], here);

			if (other[0][0] == c[i]) {
				other[0][1] = max(other[0][1], here);
			} else if (other[0][1] < here) {
				other[1][0] = other[0][0];
				other[1][1] = other[0][1];
				other[0][0] = c[i];
				other[0][1] = here;
			} else if (other[1][1] < here) {
				other[1][0] = c[i];
				other[1][1] = here;
			}
		}

		long res = max(0L, other[0][1]);
		return res;
	}

	void print(int[] a) {
		out.print(a[0]);
		for (int i = 1; i < a.length; i++) out.print(" " + a[i]);
		out.println();
	}

	class MyScanner {
		String line;
		BufferedReader reader;
		StringTokenizer tokenizer;

		public MyScanner(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public void eat() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					line = reader.readLine();
					if (line == null) {
						tokenizer = null;
						return;
					}
					tokenizer = new StringTokenizer(line);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}

		public String next() {
			eat();
			return tokenizer.nextToken();
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}

		public boolean hasNext() {
			eat();
			return (tokenizer != null && tokenizer.hasMoreElements());
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
	}
}