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.io.*;
import java.math.*;
import java.security.KeyStore.Entry;
import java.util.*;


 public class Main {
	private InputStream is;
	private PrintWriter out;
	int time = 0, dp[][], DP[][], start[], end[], val[], black[], MOD = (int)(1e9+7), arr[], arr1[];
	int MAX = 10000000, N, K, p;
	ArrayList<Integer>[] amp, amp1;
	boolean b[], b1[];
	Pair prr[];
	char ch[][];
	HashSet<Integer> hs = new HashSet<>();

	public static void main(String[] args) throws Exception {
		new Thread(null, new Runnable() {
		public void run() {
			try {

		//new Main().soln();
		} catch (Exception e) {
				System.out.println(e);
			}
		}
	}, "1", 1 << 26).start();
		new Main().soln();
	}
	void solve() {
		int n = ni();
		char ch[] =  ns().toCharArray();
		int q = ni();
		int ans[][] = new int[26][n+1];
		for(int i = 0; i<26;i++){
			char c = (char)(i+'a');
			for(int x = 1; x<=n;x++){
				//System.out.println(i+" "+x);
			int k = 0, j = 0, temp = 0, ans1 = 0;
				while(j<n && k<n){
				//	System.out.println(j+" "+k +" "+temp);
					if(ch[j]!=c){
						if(temp<x){
							temp++;
							j++;
						}
						else{
							while(k<j && temp==x){
								if(ch[k]!=c){
									temp--;
								}
								k++;
							}
						}
					}
					else j++;
					ans1 = Math.max(ans1, j-k);
				}
				ans[i][x] = ans1;
			}
		}
		for(int i = 0; i< q;i++){
			int a1 = ni();
			char c = ns().charAt(0);
			out.println(ans[c-'a'][a1]);
		}
		
	}
	boolean check(int ans[], int a[], int b[]){
		int cnt = 0;
		for(int i = 0; i< ans.length;i++){
			if(a[i] == ans[i]) cnt++;
		}
		if(cnt!=(ans.length-1)) return false;
		cnt = 0;
		for(int i = 0; i< ans.length;i++){
			if(b[i] == ans[i]) cnt++;
		}
		if(cnt!=(ans.length-1)) return false;
		return true;
	}
	int calc(int[] c, int[] d, int n, int m, int e){
		int ans = 0;
		for(int i = 0;i<n;i++){
			for(int j = 0;j<m ;j++){
				if(Math.abs(c[i]-d[j])<=e) ans++;
			}
		}
		return ans;
	}
	int binarySearch(int arr[], int key){
		int low = 0, high = arr.length-1;
		int ans = high;
		if(key>arr[high]){
			return -high-2;
		}
		while(high>=low){
			int mid = (low+high)/2;
			if(arr[mid]>=key){
				high = mid-1;
				ans = mid;
			}
			else low = mid+1;
		}
		if(arr[ans]!=key) return (-ans-1);
		return ans;
	}
	int min(int x,int y){
		if(x<y) return x;
		return y;
	}
	int max(int x,int y){
		if(x>y) return x;
		return y;
	}
	void dfs(int x){
		b[x] = true;
		for(int i : amp[x]){
			if(!b[i]){
				dfs(i);
			}
		}
	}
	void buildGraph(int m){
		while(m-->0)
		{
			int x = ni(), y = ni();
			amp[--x].add(--y);
		}
	}
	long modInverse(long a, long mOD2){
        return  power(a, mOD2-2, mOD2);
	}
	long power(long x, long y, long m)
	{
		if (y == 0)
    return 1;
	long p = power(x, y/2, m) % m;
	p = (p * p) % m;

	return (y%2 == 0)? p : (x * p) % m;
	}
	boolean isPrime(int x){
		for(int i = 2;i*1L*i<=x;i++) if(x%i==0) return false;
		return true;
	}
	public long gcd(long a, long b){
		if(b==0) return a;
		return gcd(b,a%b);
	}
	void failFn(String str, int arr[]){
		int len = 0;
		arr[0] = 0;
		int i = 1;
		while(i<arr.length){
			if(str.charAt(i)==str.charAt(len)){
				arr[i++] = ++len;
				continue;
			}
			if(len == 0){
				arr[i] = len;
				i++;
				continue;
			}
			if(str.charAt(i)!=str.charAt(len)){
				len = arr[len-1];
			}
		}
	}
	static class ST1{
		int arr[], st[], size;
		ST1(int a[]){
			arr = a.clone();
			size = 10*arr.length;
			st = new int[size];
			build(0,arr.length-1,1);
		}
		void build(int ss, int se, int si){
			if(ss==se){
				st[si] = arr[ss];
				return;
			}
			int mid = (ss+se)/2;
			int val = 2*si;
			build(ss,mid,val); build(mid+1,se,val+1);
			st[si] = (st[val]+ st[val+1]);
		}
		int get(int ss, int se, int l, int r, int si){
			if(l>se || r<ss || l>r) return Integer.MAX_VALUE;
			if(l<=ss && r>=se) return st[si];
			int mid = (ss+se)/2;
			int val = 2*si;
			return (get(ss,mid,l,r,val)+ get(mid+1,se,l,r,val+1));
		}
	}
	static class ST{
		int arr[],lazy[],n;
		ST(int a){
			n = a;
			arr = new int[10*n];
			lazy = new int[10*n];
		}
		void up(int l,int r,int val){
			update(0,n-1,0,l,r,val);
		}
		void update(int l,int r,int c,int x,int y,int val){
			if(lazy[c]!=0){
				lazy[2*c+1]+=lazy[c];
				lazy[2*c+2]+=lazy[c];
				if(l==r)
					arr[c]+=lazy[c];
				lazy[c] = 0;
			}
			if(l>r||x>y||l>y||x>r)
				return;
			if(x<=l&&y>=r){
				lazy[c]+=val;
				return ;
			}
			int mid = l+r>>1;
			update(l,mid,2*c+1,x,y,val);
			update(mid+1,r,2*c+2,x,y,val);
			arr[c] = (arr[2*c+1]+ arr[2*c+2]);
		}
		int an(int ind){
			return ans(0,n-1,0,ind);
		}
		int ans(int l,int r,int c,int ind){
			if(lazy[c]!=0){
				lazy[2*c+1]+=lazy[c];
				lazy[2*c+2]+=lazy[c];
				if(l==r)
					arr[c]+=lazy[c];
				lazy[c] = 0;
			}
			if(l==r)
				return arr[c];
			int mid = l+r>>1;
			if(mid>=ind)
				return ans(l,mid,2*c+1,ind);
			return ans(mid+1,r,2*c+2,ind);
		}
	}
	public static class FenwickTree {
		
	    int[] array; // 1-indexed array, In this array We save cumulative information to perform efficient range queries and updates

	    public FenwickTree(int size) {
	        array = new int[size + 1];
	    }
	    public int rsq(int ind) {
	        assert ind > 0;
	        int sum = 0;
	        while (ind > 0) {
	            sum += array[ind];
	            //Extracting the portion up to the first significant one of the binary representation of 'ind' and decrementing ind by that number
	            ind -= ind & (-ind);
	        }
	        return sum;
	    }
	    public int rsq(int a, int b) {
	        assert b >= a && a > 0 && b > 0;
	        return rsq(b) - rsq(a - 1);
	    }
	    public void update(int ind, int value) {
	        assert ind > 0;
	        while (ind < array.length) {
	            array[ind] += value;
	            //Extracting the portion up to the first significant one of the binary representation of 'ind' and incrementing ind by that number
	            ind += ind & (-ind);
	        }
	    }
	    public int size() {
	        return array.length - 1;
	    }
	}
	public static int[] shuffle(int[] a, Random gen){
		for(int i = 0, n = a.length;i < n;i++)
		{ 
			int ind = gen.nextInt(n-i)+i; 
			int d = a[i]; 
			a[i] = a[ind];
			a[ind] = d; 
		} 
	return a; 
	}

	class Pair implements Comparable<Pair>{
		int u, v;
		Pair(int u, int v){
			this.u = u;
			this.v = v;
		}public int hashCode() {
			return Objects.hash();
		}
		public boolean equals(Object o) {
			Pair other = (Pair) o;
			return ((u == other.u && v == other.v));// || (v == other.u && u == other.v));
		}
		public int compareTo(Pair other) {
			//return Integer.compare(val, other.val);
			return Long.compare(u, other.u) != 0 ? (Long.compare(u, other.u)) : (Long.compare(other.v,v));
		}	
		public String toString() {
			return "[u=" + u + ", v=" + v + "]";
		}
	}
	long power(long x, long y, int mod){
		long ans = 1;
		while(y>0){
			if(y%2==0){
				x = (x*x)%mod;
				y/=2;
			}
			else{
				ans = (x*ans)%mod;
				y--;
			}
		}
		return ans;
	}
	void soln() {
		is = System.in;
		out = new PrintWriter(System.out);
		long s = System.currentTimeMillis();
		solve();
		//out.close();
		out.flush();
		//tr(System.currentTimeMillis() - s + "ms");
	}


	// To Get Input
	// Some Buffer Methods
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}
	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
									// ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n && !(isSpaceChar(b))) {
			buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;

	private void tr(Object... o) {
		if (!oj)
			System.out.println(Arrays.deepToString(o));
	}
}