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.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.*;

public class C418 {

  private static final String REGEX = " ";
  private static final Boolean DEBUG = false;
  private static final String FILE_NAME = "input.txt";


  public static void main(String[] args) throws IOException {
    Solver solver = new Solver();
    solver.readData();
    solver.solveAndPrint();
  }

  private static class Solver {
    int n;
    char s[];
    int q;

    char c[];
    int m[];


    void readData() throws IOException {
      InputStream in = DEBUG ? new FileInputStream(FILE_NAME) : System.in;
      Scanner scanner = new Scanner(in);
      n = scanner.nextInt();
      scanner.nextLine();
      String s = scanner.nextLine();
      this.s = new char[n];
      for (int i = 0; i < n; i++) {
        this.s[i] = s.charAt(i);
      }
      q = scanner.nextInt();
      m = new int[q];
      c = new char[q];
      scanner.nextLine();
      for (int i = 0; i < q; i++) {
        String[] split = scanner.nextLine().split(" ", 2);
        m[i] = Integer.parseInt(split[0]);
        c[i] = split[1].charAt(0);
      }
    }

    void solveAndPrint() {

      Map<Character, Map<Integer, Integer>> map = new HashMap<>();
      for (char c = 'a'; c <= 'z'; c++) {
        map.putIfAbsent(c, new HashMap<>());
        for (int curI = 1; curI <= n; curI++) {
          int startIndex = 0;
          int endIndex = 0;
          int maxLength = 0;
          int counter = curI;
          Queue<Integer> nextStart = new ArrayDeque<>();
          for (int i = 0; i < n; i++) {
            if (s[i] == c) {
              endIndex++;
            } else {
              counter--;
              nextStart.add(i);
              if (counter == -1) {
                maxLength = max(maxLength, endIndex - startIndex);
                counter++;
                startIndex = nextStart.remove() + 1;
              }
              endIndex++;
            }
          }
          maxLength = max(maxLength, endIndex - startIndex);
          map.get(c).put(curI, maxLength);
        }
      }


      StringBuilder result = new StringBuilder();
      for (int j = 0; j < q; j++) {
        result.append(map.get(c[j]).get(m[j]));
        result.append("\n");
      }
      System.out.println(result);

    }


    @SuppressWarnings("SameParameterValue")
    int[] splitInteger(String string, int n) {
      final String[] split = string.split(REGEX, n);
      int[] result = new int[split.length];
      for (int i = 0; i < n; ++i) {
        result[i] = Integer.parseInt(split[i]);
      }
      return result;
    }

    public int[] splitInteger(String string) {
      return splitInteger(string, 0);
    }


    @SuppressWarnings("SameParameterValue")
    long[] splitLong(String string, int n) {
      final String[] split = string.split(REGEX, n);
      long[] result = new long[split.length];
      for (int i = 0; i < n; ++i) {
        result[i] = Long.parseLong(split[i]);
      }
      return result;
    }

    public long[] splitLong(String string) {
      return splitLong(string, 0);
    }

    @SuppressWarnings("SameParameterValue")
    double[] splitDouble(String string, int n) {
      final String[] split = string.split(REGEX, n);
      double[] result = new double[split.length];
      for (int i = 0; i < n; ++i) {
        result[i] = Double.parseDouble(split[i]);
      }
      return result;
    }

    public double[] splitDouble(String string) {
      return splitDouble(string, 0);
    }

    @SuppressWarnings("SameParameterValue")
    String[] splitString(String string, int n) {
      return string.split(REGEX, n);
    }

    public String[] splitString(String string) {
      return splitString(string, 0);
    }

    public int max(int a, int b) {
      return Math.max(a, b);
    }

    public long max(long a, long b) {
      return Math.max(a, b);
    }

    public int min(int a, int b) {
      return Math.min(a, b);
    }

    public long min(long a, long b) {
      return Math.min(a, b);
    }

    public double max(double a, double b) {
      return Math.max(a, b);
    }

    public double min(double a, double b) {
      return Math.min(a, b);
    }

    private final static int MOD = 1000000009;

    int multMod(int a, int b) {
      return ((a % MOD) * (b % MOD)) % MOD;
    }

    int sumMod(int a, int b) {
      return ((a % MOD) + (b % MOD)) % MOD;
    }

    long multMod(long a, long b) {
      return ((a % MOD) * (b % MOD)) % MOD;
    }

    long sumMod(long a, long b) {
      return ((a % MOD) + (b % MOD)) % MOD;
    }
  }


}
