Description

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

Examples output

3
6
5

題意

給定一個字串，有 q$q$ 次查詢，每次查詢告訴你最喜歡的字元以及將其他字元替換為它的次數，問字串中連續最喜歡的字元最大有多長。

思路

dp[i][j]$dp[i][j]$ 代表 i$i$ 這一個字元在可以替換 j$j$ 次的條件下所能組成的最大長度。

然後我們便可以在 O(n2)$O(n^2)$ 的時間複雜度下打表出所有的情況，根據查詢一一輸出即可。

當然這道題尺取法也可行，時間複雜度 O(n×q)$O(n\times q)$ 。

AC 程式碼

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);
using namespace std;
const int maxn = 1e5+10;
const int mod = 1e9+7;
typedef long long LL;

char str[maxn];
int dp[26][maxn],n;

void init()
{
    for(int ch=0; ch<26; ch++)
    {
        for(int i=0; i<n; i++)
        {
            int now = 0;
            for(int j=i; j<n; j++)
            {
                if(str[j]!='a'+ch)
                    now++;
                dp[ch][now] = max(dp[ch][now],j-i+1);
            }
        }
        for(int i=1; i<=n; i++)
            dp[ch][i] = max(dp[ch][i],dp[ch][i-1]);
    }
}

int main()
{
    int q;
    cin>>n>>str>>q;
    init();
    while(q--)
    {
        int m;
        char c;
        cin>>m>>c;
        cout<<dp[c-'a'][m]<<endl;
    }
    return 0;
}