Codeforces Round #418 (Div. 2) C. An impassioned circulation of affection

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在Nadeko的生日即将来临之际，她决定为庆祝派对装饰房间，制作了一条由Dianthus形状纸片组成的长花环。为了进一步美化花环并猜测兄长Koyomi的喜好，Nadeko计划通过重新上色部分纸片来提升花环的'Koyomity'，即找出含有Koyomi最爱颜色的最长连续片段。本文探讨了如何在有限的修改次数下，通过算法寻找最优解。

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

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.

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 1590
#define N 26
#define MOD 1000000
#define INF 1000000009
const double eps = 1e-6;
const double PI = acos(-1.0);
/*
一开始的思路：线段相连，一直求相连的长度最长的两个线段增加的c字符数目，减少m的次数，一直到无法增加
编程复杂，时间复杂度高
题解：
首先 目的是让要求的字符c的连续串长度最长，那么我们应该在一个c的连续串附近交换字符，否则无法增大最优解
对于每一个字符c和长度，枚举出需要交换m个c字符的时候最长连续序列的长度然后直接查询！
*/
int ans[N][MAXN], l ,n;
char s[MAXN];
int main()
{
    scanf("%d", &l);
    scanf("%s", s);
    for (int c = 0; c < N; c++)
    {
        for (int i = 0; i < l; i++)
        {
            int dif = 0;
            for (int j = i; j < l; j++)
            {
                if (s[j] != c + 'a')
                    dif++;
                ans[c][dif] = max(ans[c][dif], j - i + 1);
            }
        }
        for (int i = 1; i <= l; i++)
            ans[c][i] = max(ans[c][i], ans[c][i - 1]);
    }
    scanf("%d", &n);
    int m;
    char c;
    while (n--)
    {
        scanf("%d %c", &m, &c);
        m = min(l, m);
        printf("%d\n", ans[c-'a'][m]);
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/6964626.html