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. 
/* Author: Ronak Agarwal */
import java.io.* ; import java.util.* ; import java.math.* ;
import static java.lang.Math.min ;
import static java.lang.Math.max ;
import static java.lang.Math.abs ;
/* Thread is created here to increase the stack size of the java code so that recursive dfs can be performed */
public class Codeshefcode{
	public static void main(String[] args) throws IOException{
		new Thread(null,new Runnable(){	public void run(){ exec_Code() ;} },"Solver",1l<<27).start() ;
	}
	static void exec_Code(){
		try{
			Solver Machine = new Solver() ;
			Machine.Solve() ; Machine.Finish() ;}
		catch(Exception e){ e.printStackTrace() ; print_and_exit() ;}
		catch(Error e){ e.printStackTrace() ; print_and_exit() ; }
	}
	static void print_and_exit(){ System.out.flush() ; System.exit(-1) ;}
}
class Reader{
	BufferedReader reader;
	StringTokenizer tkz;
	Reader(InputStream input) {
		reader = new BufferedReader(new InputStreamReader(input)) ;
		tkz = new StringTokenizer("") ;
	}
	String s() throws IOException{
		while (!tkz.hasMoreTokens()) tkz = new StringTokenizer(reader.readLine()) ;
		return tkz.nextToken() ;
	}
	int i() throws IOException{ return Integer.parseInt(s()) ;}
	long l() throws IOException{ return Long.parseLong(s()) ;}
	double d() throws IOException{	return Double.parseDouble(s()) ;}
}
/* All the useful functions,constants,renamings are here*/
class Template{
	/* Constants Section */
	final int INT_MIN = Integer.MIN_VALUE ; final int INT_MAX = Integer.MAX_VALUE ;
	final long LONG_MIN = Long.MIN_VALUE ; final long LONG_MAX = Long.MAX_VALUE ;
	static long MOD = 1000000007 ;
	static Reader ip = new Reader(System.in) ;	
	static PrintWriter op = new PrintWriter(System.out) ;
	/* Methods for writing */
	static void p(Object o){ op.print(o) ; }
	static void pln(Object o){ op.println(o) ;}
	static void Finish(){ op.flush();	op.close(); }
	/* Implementation of operations modulo MOD */
	static long inv(long a){ return powM(a,MOD-2) ; }
	static long m(long a,long b){ return (a*b)%MOD ; }
	static long d(long a,long b){ return (a*inv(b))%MOD ; }
	static long powM(long a,long b){
		if(b<0) return powM(inv(a),-b) ;
		long val=a ; long ans=1 ;
		while(b!=0){
			if((b&1)==1) ans = (ans*val)%MOD ; 
			val = (val*val)%MOD ; 
			b/=2 ;
		}
		return ans ;
	}
	/* Renaming of some generic utility classes */
	final static class mylist extends ArrayList<Integer>{}
	final static class myset extends TreeSet<Integer>{}
	final static class mystack extends Stack<Integer>{}
	final static class mymap extends TreeMap<Long,Integer>{}
}
/* Implementation of the pair class, useful for every other problem */
class pair implements Comparable<pair>{
	long x ; long y ;
	pair(long _x,long _y){ x=_x ; y=_y ;} 
	public int compareTo(pair p){
		return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
	}
}
/* Main code starts here */
class Solver extends Template{
	int res[][] ;
	public void Solve() throws IOException{
		int n = ip.i() ;
		res = new int[26][n+1] ;
		for(int i=0 ; i<26 ; i++) Arrays.fill(res[i],INT_MIN) ;
		char c[] = ip.s().toCharArray() ;
		int arr[] = new int[n+1] ;
		for(int i=1 ; i<=n ; i++) arr[i] = c[i-1]-'a' ;
		for(int k=0 ; k<26 ; k++){
			int dp[][] = new int[n+1][n+1] ;
			for(int i=1 ; i<=n ; i++){
				for(int j=0 ; j<=n ; j++){
					dp[i][j] = (arr[i]==k ? dp[i-1][j] : (j==0 ? -1 : dp[i-1][j-1]))+1 ;
				}
			}
			for(int i=1 ; i<=n ; i++)
				for(int j=0 ; j<=n ; j++)
					res[k][j] = max(res[k][j],dp[i][j]) ;
		}
		int q = ip.i() ;
		while(q--!=0){
			int mi = ip.i() ;
			int ci = (ip.s().charAt(0)-'a') ;
			pln(res[ci][mi]) ;
		}
	}
}
