Codeforces 814C An impassioned circulation of affection【Dp】

C. An impassioned circulation of affection

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 hasn pieces numbered from 1 to n from left to right, and the i-th piece has a colour s_i, 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 theKoyomity 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 of3. Thus the Koyomity of this garland equals3.

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 hasq plans on this, each of which can be expressed as a pair of an integerm_i and a lowercase letterc_i, meanings of which are explained above. You are to find out the maximumKoyomity 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 letterss₁s₂...s_n 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: thei-th among them contains an integer m_i (1 ≤ m_i ≤ n) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letterc_i — Koyomi's possible favourite colour.

Output

Output q lines: for each work plan, output one line containing an integer — the largestKoyomity achievable after repainting the garland according to it.

Examples

Input

6koyomi31 o4 o4 m

Output

365

Input

15yamatonadeshiko101 a2 a3 a4 a5 a1 b2 b3 b4 b5 b

Output

3457812345

Input

10aaaaaaaaaa210 b10 z

Output

1010

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 aKoyomity of 6;
• In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in aKoyomity of 5. 

题目大意：

给你一个长度为N的字符串，然后给出q个查询，每个查询两个元素，数字表示可以改动的字符数量，字符表示查询改动后的最长连续当前字符的长度。

思路：

考虑到一共只有26个字母，如果我们O（n*q）的去Dp的话 ，会TLE（最坏差不多3e8的操作）掉；

所以我们考虑预处理Dp.设定dp【i】【j】【k】表示字母为i，到了位子j，用了k次改动机会的最长字母i串长度。

那么其状态转移方程不难写出：

时间复杂度O（26*n*n）；

然后我们再维护一个结果即可。

可以做到O（1）查询。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;char a[1550];int ans[26][1550];int dp[27][1550][1550];int main(){    int n;    while(~scanf("%d",&n))    {        scanf("%s",a+1);        memset(ans,0,sizeof(ans));        memset(dp,0,sizeof(dp));        for(int k=0;k<26;k++)        {            for(int i=1;i<=n;i++)            {                for(int j=0;j<=i;j++)                {                    if(a[i]-'a'==k)                    {                        dp[k][i][j]=1;                        dp[k][i][j]=max(dp[k][i][j],dp[k][i-1][j]+1);                        if(j-1>=0)dp[k][i][j]=max(dp[k][i][j],dp[k][i-1][j-1]+1);                    }                    else                    {                        dp[k][i][j]=0;                        if(j-1>=0)dp[k][i][j]=max(dp[k][i][j],dp[k][i-1][j-1]+1);                    }                }            }        }        for(int i=0;i<26;i++)        {            for(int j=0;j<=n;j++)            {                for(int k=1;k<=n;k++)                {                    ans[i][j]=max(ans[i][j],dp[i][k][j]);                }            }        }        int q;        scanf("%d",&q);        while(q--)        {            char s[15];            int qq;            scanf("%d%s",&qq,s);            printf("%d\n",ans[s[0]-'a'][qq]);        }    }}