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.util.*;
public class ImpassionedCirculation {

	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO Auto-generated method stub
		
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
		int n = Integer.parseInt(bf.readLine());
		String line = bf.readLine();
		int q = Integer.parseInt(bf.readLine());
		
		String alph = "abcdefghijklmnopqrstuvwxyz";
		
		int[][] ans = new int[26][n+1];
		for (int i = 0; i < 26; i++) {
			for (int j = 0; j <= n; j++) {
				ans[i][j] = j;
			}
		}
		
		for (int i = 0; i < 26; i++) {
			
			
			char c = alph.charAt(i); //examining current character as if favorite
			ArrayList<block> bl = new ArrayList<block>(); //creating list of runs
			
			boolean inRun = false; //whether or not the sequence is in a block (starts off as not)
			int start = 0; //start of a current block (means nothing here)
			
			for (int j = 0; j < n; j++) {
				
				if (line.charAt(j)==c) {
					if (!inRun) { //start a block
						inRun = true;
						start = j;
					}
				}
				else {
					if (inRun) {
						inRun = false;
						bl.add(new block(start, j-1)); //block ended at previous character
					}
				}
			}
			if (inRun) { //add a block for the last section
				bl.add(new block(start, n-1));
			}
			
			//calculated runs
			
			for (int j = 0; j < bl.size(); j++) {
				//start at this block
				
				block st = bl.get(j);
				int cursize = st.size();
				int cost = 0;
				int prevh = st.high;
				ans[i][cost] = Math.max(ans[i][cost], cursize);
				
				for (int k = j+1; k < bl.size(); k++) {
					
					block b = bl.get(k);
					int ccost = b.low-prevh-1; //cost of going between the blocks
					
					for (int l = 0; l <= ccost; l++) {
						ans[i][l+cost] = Math.max(ans[i][l+cost], cursize+l);
					}
					
					cost+=ccost; //this adds the cost of appending the next item to the current cost
					
					cursize += b.size(); // adds size of next block
					
					cursize+=ccost; //this adds the size we appended
					
					prevh = b.high;//update previous high 
					
					ans[i][cost] = Math.max(ans[i][cost], cursize); //update current position just in case
				}
				for (int k = cost; k <= n; k++) { //run with appending to open ends (extra check / add on for end or beginning)
					int cur = cursize+(k-cost);
					cur = Math.min(cur, n);
					ans[i][k] = Math.max(ans[i][k], cur);
				}
//				if (i == 0) System.out.println(j + " " + Arrays.toString(ans[0]));

			}
		}
		
		//double checking ...
		for (int i = 0; i < 26; i++) {
			for (int j = 1; j <= n; j++) {
				ans[i][j] = Math.max(ans[i][j], ans[i][j-1]);
			}
		}
		//everything should be calculated??
		
		for (int i = 0; i < q; i++) {
			String[] cs = bf.readLine().split(" ");
			int num = Integer.parseInt(cs[0]);
			int index = alph.indexOf(cs[1].charAt(0));
			pw.println(ans[index][num]);
		}
//		System.out.println(Arrays.toString(ans[0]));
		pw.close();
	}
	
	static class block  {
		public int low, high;
		public block (int low, int high) {
			this.low = low;
			this.high = high;
		}
		
		public int size() {
			return (high-low)+1;
		}
		
		public String toString() {
			return "{" + low + ", " + high + "}";
		}
	}

}
