Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!

Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from left to right, and the i-th piece has a colour si, denoted by a lowercase English letter. Nadeko will repaint at most m of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c — Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.

For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o". Among all subsegments containing pieces of "o" only, "ooo" is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.

But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has q plans on this, each of which can be expressed as a pair of an integer mi and a lowercase letter ci, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.

Input

The first line of input contains a positive integer n (1 ≤ n ≤ 1 500) — the length of the garland.

The second line contains n lowercase English letters s1s2... sn as a string — the initial colours of paper pieces on the garland.

The third line contains a positive integer q (1 ≤ q ≤ 200 000) — the number of plans Nadeko has.

The next q lines describe one plan each: the i-th among them contains an integer mi (1 ≤ mi ≤ n) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci — Koyomi's possible favourite colour.

Output

Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.

Examples

Input

6
koyomi
3
1 o
4 o
4 m


Output

3
6
5


Input

15
yamatonadeshiko
10
1 a
2 a
3 a
4 a
5 a
1 b
2 b
3 b
4 b
5 b


Output

3
4
5
7
8
1
2
3
4
5


Input

10
aaaaaaaaaa
2
10 b
10 z


Output

10
10

Note

In the first sample, there are three plans: 

  * In the first plan, at most 1 piece can be repainted. Repainting the "y" piece to become "o" results in "kooomi", whose Koyomity of 3 is the best achievable; 
  * In the second plan, at most 4 pieces can be repainted, and "oooooo" results in a Koyomity of 6; 
  * In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in a Koyomity of 5. 
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{       static PrintWriter out=new PrintWriter(System.out);static FastScanner in = new FastScanner(System.in);static class FastScanner {BufferedReader br;StringTokenizer stok;FastScanner(InputStream is) {br = new BufferedReader(new InputStreamReader(is));}
        String next() throws IOException {while (stok == null || !stok.hasMoreTokens()) {String s = br.readLine();if (s == null) {return null;}
                stok = new StringTokenizer(s);}return stok.nextToken();}
        int ni() throws IOException {return Integer.parseInt(next());}long nl() throws IOException {return Long.parseLong(next());}double nd() throws IOException {return Double.parseDouble(next());}char nc() throws IOException {return (char) (br.read());}String ns() throws IOException {return br.readLine();}
        int[] nia(int n) throws IOException{int a[] = new int[n];for (int i = 0; i < n; i++)a[i] = ni();return a;}long[] nla(int n) throws IOException {long a[] = new long[n];for (int i = 0; i < n; i++)a[i] = nl();return a;}
        double[] nda(int n)throws IOException {double a[] = new double[n];for (int i = 0; i < n; i++)a[i] = nd();return a;}int [][] imat(int n,int m) throws IOException{int mat[][]=new int[n][m];for(int i=0;i<n;i++){for(int j=0;j<m;j++)mat[i][j]=ni();}return mat;}
    }
	static long mod=Long.MAX_VALUE;
	static int n;
	static String s;
	static int ans[][];
	public static void main (String[] args) throws java.lang.Exception
	{   int i,j;
	    //HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
		/*  if(hm.containsKey(z))
		        hm.put(z,hm.get(z)+1);
		    else
		        hm.put(z,1);
		 */       
        n=in.ni();
        s=in.next();
        ans=new int[n][26];
        for(i=0;i<26;i++)
        {   char cc=(char)(i+'a');
            func3(cc);
        }
        int xx=in.ni();
        for(i=0;i<xx;i++)
        {   int m=in.ni();
            char x=in.next().charAt(0);
            int p=x-'a';
            if(m==n)
                out.println(m);
            else    
                out.println(ans[m][p]);
            //func(0,0,m,x);
            //func3(x);
            //out.println(ans);
        }
        
        
        out.close();
	}
	static void func3(char c)
	{   int i,j,m;
	    int k=c-'a';
	    int dp[][]=new int[n][n];
	    if(s.charAt(0)==c)
	        dp[0][0]=1;
	    for(i=1;i<n;i++)
	        dp[0][i]=1;
	    for(i=1;i<n;i++)
	    {   for(j=0;j<n;j++)
	        {
	            if(s.charAt(i)==c)
	                dp[i][j]=dp[i-1][j]+1;
	            else if(j>0)
	                dp[i][j]=dp[i-1][j-1]+1;
	        }
	    }   
	    int max=-1;
	        for(i=0;i<n;i++)
            {   for(j=0;j<n;j++)
                    max=Math.max(max,dp[j][i]);
                ans[i][k]=max;
            }
	    
	    
	    //for(i=0;i<=n;i++)
	    //{   for(j=0;j<=n;j++)
	    //        out.print(dp[i][j][k]+" ");
	    //    out.println();     
	    //}
	}
	static class pair implements Comparable<pair>{
		int x, y;
		public pair(int x, int y){this.x = x; this.y = y;}
		@Override
		public int compareTo(pair arg0) 
		{   if(x<arg0.x)    return -1;
		    else if(x==arg0.x)
		    {   if(y<arg0.y)    return -1;
		        else if(y>arg0.y)   return 1;
		        else    return 0;
		    }
		    else    return 1;
		}
	}
	static long gcd(long a,long b)
	{   if(b==0)
	        return a;
	    return gcd(b,a%b);    
	}
	static long exponent(long a,long n)
	{   long ans=1;
	    while(n!=0)
	    {   if(n%2==1)
	            ans=(ans*a)%mod;
	       a=(a*a)%mod;
	       n=n>>1;
	    }
	    return ans;
	}
	static int binarySearch(int a[], int item, int low, int high) 
    {   if (high <= low) 
            return (item > a[low])?  (low + 1): low; 
        int mid = (low + high)/2; 
        if(item == a[mid]) 
            return mid+1; 
        if(item > a[mid]) 
            return binarySearch(a, item, mid+1, high); 
        return binarySearch(a, item, low, mid-1); 
    } 
    static void merge(int arr[], int l, int m, int r) {   
        int n1 = m - l + 1; int n2 = r - m; int L[] = new int [n1]; int R[] = new int [n2]; 
        
        for (int i=0; i<n1; ++i) L[i] = arr[l + i]; for (int j=0; j<n2; ++j) R[j] = arr[m + 1+ j]; int i = 0, j = 0; int k = l; while (i < n1 && j < n2) {   if (L[i] <= R[j]) {   arr[k] = L[i]; i++; } else{   arr[k] = R[j]; j++; } k++; } while (i < n1){   arr[k] = L[i]; i++; k++; } while (j < n2) { arr[k] = R[j];   j++;     k++; } 
    } 
    static void Sort(int arr[], int l, int r) {if (l < r) {   int m = (l+r)/2; Sort(arr, l, m); Sort(arr , m+1, r); merge(arr, l, m, r); } } 
     static void sort(int a[])
    {Sort(a,0,a.length-1);}
}