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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Map.Entry;
import java.util.TreeMap;
public class C264 
{
    static class SegmentTree
    {
        int n;
        long t[];
        SegmentTree(int n)
        {
            this.n=n;
            t=new long[2*n+10];
        }
        void init()
        {
            Arrays.fill(t,0);
        }
        void update(int index,long value)
        {
            index+=n;
            for(t[index]=value;index>1;index>>=1)
                t[index>>1]=Math.max(t[index],t[index^1]);
        }
        long query(int l,int r)
        {
            long res=0;
            r++;
            for(l+=n,r+=n;l<r;l>>=1,r>>=1)
            {
                if((l&1)>0)
                    res=Math.max(res,t[l++]);
                if((r&1)>0)
                    res=Math.max(res,t[--r]);
            }
            return res;
        }
    }
    static class Scanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Scanner() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Scanner(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException {
            byte[] buf = new byte[10000]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }
        int[] nextIntArray(int n) throws IOException
        {
            int a[]=new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();
            return a;
        }
        long[] nextLongArray(int n) throws IOException
        {
            long a[]=new long[n];
            for(int i=0;i<n;i++)
                a[i]=nextLong();
            return a;
        }
    }
    
    static class MultiSet extends TreeMap<Long,Integer>
    {
        void insert(long x)
        {
            int c=getOrDefault(x,0);
            c++;
            put(x,c);
        }
        int count(long x)
        {
            return getOrDefault(x,0);
        }
        void remove(long x)
        {
            int c=get(x);
            c--;
            if(c!=0)
                put(x,c);
            else
                super.remove(x);
        }
        
    }
    public static void main(String args[]) throws IOException
    {
        Scanner in=new Scanner();
        PrintWriter out=new PrintWriter(System.out);
        int n=in.nextInt();
        int q=in.nextInt();
        long v[]=in.nextLongArray(n);
        int c[]=in.nextIntArray(n);
        long val[]=new long[n+1];
        while(q--!=0)
        {
            long st[][]=new long[2][2];
            st[0][1]=st[1][1]=0;
            Arrays.fill(val,Long.MIN_VALUE);
            long a=in.nextLong();
            long b=in.nextLong();
            long ans=0;
            for(int i=0;i<n;i++)
            {
                long x=b*v[i];
                if(c[i]==st[0][0])
                    x+=st[1][1];
                else
                    x+=st[0][1];
                if(val[c[i]]!=Long.MIN_VALUE)
                    x=Math.max(x,a*v[i]+val[c[i]]);
                if(val[c[i]]<x)
                {
                    ans=Math.max(ans,x);
                    if(st[0][0]==c[i])
                        st[0][1]=x;
                    else if(st[0][1]<=x)
                    {
                        st[1][0]=st[0][0];
                        st[1][1]=st[0][1];
                        st[0][0]=c[i];
                        st[0][1]=x;
                    }
                    else if(st[1][1]<x)
                    {
                        st[1][0]=c[i];
                        st[1][1]=x;
                    }
                                
                    val[c[i]]=x;
                }
            }
            
            out.println(ans);
        }
        out.close();
    }

}