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.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
       int x,y;
        Student(int x,int y){
            this.x=x;
            this.y=y;
            //this.z=z;
        }
    }
    static int prime[];
    static void sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        prime= new int[n+1]; 
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == 0)
            { 
                // Update all multiples of p
                prime[p]=p;
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = p; 
            } 
        } 
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        return c.x-b.x;
    } 
} 
static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
 
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    static class Trie{
        HashMap<Character,Trie>map;
         int c;
        Trie(){
            map=new HashMap<>();
            //z=null;
            //o=null;
            c=0;
        }
    }
    //static long ans;
    static int parent[];
    static int rank[];
    static int b[][];
    static int bo[];
    static int ho[];
    static int seg[];
    //static int pos;
   // static long mod=1000000007;
    //static int dp[][];
    static HashMap<Integer,Integer>map;
    static PriorityQueue<Student>q=new PriorityQueue<>();
    //static Stack<Integer>st;
   // static ArrayList<Character>ans;
    static ArrayList<ArrayList<Integer>>adj;
    static long ans;
    static Trie root;
    static long fac[];
    static long mod=(long)(1000000007);
    //static int ans;
    static void solve()throws IOException{
        FastReader sc=new FastReader();
        int n,i,j,k,l,c,ans,x,y;
        String s;
        StringBuilder sb=new StringBuilder();
        n=sc.nextInt();
        s=sc.nextLine();
        int dp[][]=new int[26][n+1];
        for(i=0;i<26;i++){
            for(j=1;j<=n;j++){
                l=0;
                c=0;
                ans=0;
                for(k=0;k<n;k++){
                    if((int)s.charAt(k)-97!=i)
                        ++c;
                    while(c>j){
                        if((int)s.charAt(l)-97!=i)
                            --c;
                        ++l;
                    }
                    ans=Math.max(ans,k-l+1);
                }
                dp[i][j]=ans;
            }
        }
        n=sc.nextInt();
        for(i=0;i<n;i++){
            s=sc.nextLine();
            String st[]=s.split(" ");
            sb.append(dp[st[1].charAt(0)-97][Integer.parseInt(st[0])]+"\n");
        }
        System.out.println(sb.toString());
    }
    static void build(StringBuilder sb,int n,char a){
        if(n==0)
        {
            sb.append(a);
            return;
        }
        int x=(int)Math.sqrt(2*n);
        if(x*(x+1)>2*n)
            x-=1;
        for(int i=0;i<=x;i++)
            sb.append(a);
        n-=(x*(x+1))/2;
        build(sb,n,(char)((int)a+1));
    }
    static void dfs(Trie root,int k){
        if(root==null)
            return;
        for(char i='A';i<='Z';i++){
            if(root.map.containsKey(i)){
                ans+=root.map.get(i).c/k;
                dfs(root.map.get(i),k);
            }
        }
    }
    static long nCr(long n, long r, 
                                    long p) 
    { 
        return (fac[(int)n]* modInverse(fac[(int)r], p) 
                % p * modInverse(fac[(int)(n-r)], p) 
                                    % p) % p; 
    } 
    //static int prime[];
    //static int dp[];
    public static void main(String[] args){
        //long sum=0;
        try {
                codeforces.solve();
                } catch (Exception e) {
                    e.printStackTrace();
                }
    }
       static long modInverse(long n, long p) 
    { 
        return power(n, p-(long)2,p); 
    } 
 static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y %(long)2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p;
 }
    /*public static long power(long x,long a) {
		if(a==0) return 1;
		if(a%2==1)return (x*1L*power(x,a-1))%mod;
		return (power((int)((x*1L*x)%mod),a/2))%mod;
	}*/
   static int find(int x) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set,
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x), yRoot = find(y); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    } 
    static long gcd(long a, long b) 
    { 
        if (a == 0){ 
             //ans+=b;
            return b;
        }
        return gcd(b % a, a); 
    }
}