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

題意：把一個區間裡面最多m個字母修改，問目標字母最長的連續區間是多少。不得不說上海全能王蔡大佬是真毒，讀錯題血崩，以為是可以不連續的，狂寫dpA不了。連續區間就很好寫了，很容易想到用一手尺取。演算法複雜度是O（n），那就很僵了，加上查詢超過1e8了常數還有點大。那我們分析一下題目，這麼多查詢裡面必定是有重複的，長度為n的序列對每個字母答案會出現不同的最大數字就是n。也就是說我如果可以修改超過n個對於答案來說也是沒有區別的。所以記憶化一下其實是O(n^2*26)
這樣就可以了。

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<map>
#include<set>
using namespace std;

//thanks to pyf ...
//thanks to qhl ...

#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define mp(x,y) make_pair(x,y)
typedef pair<int, int> PII;
typedef long long ll;

const int N = 1e6 + 5;

int ans[1505][30];

int rular(int m, char x, string s)
{
    int l = -1, r = 0;
    int cur_used = 0;
    int cur_Max = 0;
    int cur_ans = 0;
    while (l <= r && r < s.length())
    {
        if (s[r] == x)
        {
            cur_ans++;
            cur_Max = max(cur_ans, cur_Max);
            r++;
        }
        else
        {
            if (cur_used < m)
            {
                cur_used++;
                cur_ans ++;
                cur_Max = max(cur_ans, cur_Max);
                r++;
            }
            else
            {
                l++;
                if (s[l] != x)
                {
                    cur_used -- ;
                }
                cur_ans -- ;
            }
        }
    }
    return ans[m][x - 'a'] = cur_Max;
}
int main()
{
    int n;
    while (cin >> n)
    {
        CLR(ans, 0);
        string s;
        cin >> s;
        int m;
        cin >> m;
        for (int i = 0; i < m; i++)
        {
            int l;
            char x;
            cin >> l >> x;
            if (!ans[l][x - 'a'])
                cout << rular(l, x, s) << endl;
            else
                cout << ans[l][x - 'a'] << endl;
        }
    }
}