C. An impassioned circulation of affection（DP+two points）Codeforces Round #418 (Div. 2)

题目链接：http://codeforces.com/contest/814/problem/C

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 has n 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 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 m_i and a lowercase letter c_i, 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... 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: the i-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 letter c_i — 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.

题意：给出一个长度为 n 的小写字母构成的字符串，询问 q 次每当替换了 m 个非 c 字符后最大的连续 c 子串的长度。（c表示Koyomi可能喜欢的颜色）

分析：时间复杂度n*n*26

　　　ans[k][d]表示当关键颜色为 k+'a' 时，替换d个字符的连续子串的最大长度。利用尺取法的思想循环得出数组。

 

 

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    char s[1505];
    while(cin>>n)
    {
        cin>>s;
        int ans[30][1505];
        for(int k=0;k<26;k++)
        {
            for(int i=0;i<n;i++)
            {
                int d=0;
                for(int j=i;j<n;j++)
                {
                    if(s[j]!='a'+k)//s[j]非 c 时要替换成 c
                        d++;
                    ans[k][d]=max(ans[k][d],j-i+1);//从i到j替换了d次能连续的最大长度
                }
            }
            for(int t=1;t<=n;t++) //在循环过程中，只考虑了之后的连续，之前的没有加上。
            {
                ans[k][t]=max(ans[k][t],ans[k][t-1]); //例如替换两个能达到的长度，那么替换三个也能达到，但是未加入计算，以此类推
            }
        }
        int num,cnt;
        char c;
        cin>>num;
        for(int i=0;i<num;i++)
        {
            cin>>cnt>>c;
            cout<<ans[c-'a'][cnt]<<endl;
        }
    }
    return 0;
}