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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 1to n from left to right, and the i-th piece has a colour si, denoted by a lowercase English letter. Nadeko will repaintat 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 Koyomityof 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 Koyomityachievable after repainting the garland according to each plan.
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 q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.
6
koyomi
3
1 o
4 o
4 m
3
6
5
15
yamatonadeshiko
10
1 a
2 a
3 a
4 a
5 a
1 b
2 b
3 b
4 b
5 b
3
4
5
7
8
1
2
3
4
5
10
aaaaaaaaaa
2
10 b
10 z
10
10
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.
题目大意：
q组询问
每组一个x和一个字符ch 表示可以替换x个字符成为ch 问最大的连续为ch的字符串长度
题解：
预处理出：F[i][j]表示把替换掉i个可以产生连续为j的字符串的最大长度
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int N=,MAXL=;
int F[N][MAXL];char s[N];
int main()
{
int n,k=;
scanf("%d",&n);
scanf("%s",s+);
for(int ch=;ch<=;ch++)
{
for(int i=;i<=n;i++)
{
k=;
for(int j=i;j<=n;j++)
{
if(ch!=s[j]-'a')k++;
F[k][ch]=max(F[k][ch],j-i+);
}
}
for(int i=;i<=n;i++)if(F[i-][ch]>F[i][ch])F[i][ch]=F[i-][ch];
}
int m,x;char s[];
scanf("%d",&m);
while(m--)
{
scanf("%d%s",&x,s);
printf("%d\n",F[x][s[]-'a']);
}
return ;
}