Codeforces Round #116 (Div. 2, ACM-ICPC Rules)---E. Cubes

Let’s imagine that you’re playing the following simple computer game. The screen displays n lined-up cubes. Each cube is painted one of m colors. You are allowed to delete not more than k cubes (that do not necessarily go one after another). After that, the remaining cubes join together (so that the gaps are closed) and the system counts the score. The number of points you score equals to the length of the maximum sequence of cubes of the same color that follow consecutively. Write a program that determines the maximum possible number of points you can score.

Remember, you may delete no more than k any cubes. It is allowed not to delete cubes at all.

Input

The first line contains three integers n, m and k (1?≤?n?≤?2·105,?1?≤?m?≤?105,?0?≤?k?<?n). The second line contains n integers from 1 to m — the numbers of cube colors. The numbers of colors are separated by single spaces.

Output

Print the maximum possible number of points you can score.

Sample test(s)

Input

10 3 2

1 2 1 1 3 2 1 1 2 2

Output

4

Input

10 2 2

1 2 1 2 1 1 2 1 1 2

Output

5

Input

3 1 2

1 1 1

Output

3

Note

In the first sample you should delete the fifth and the sixth cubes.

In the second sample you should delete the fourth and the seventh cubes.

In the third sample you shouldn’t delete any cubes.

对颜色分组，然后组内用尺取法就行了

/*************************************************************************
    > File Name: CF-116-E.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年03月30日 星期一 19时53分05秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 200100;
vector <int> cols[100100];

int main()
{
    int n, m, k;
    while (~scanf("%d%d%d", &n, &m, &k))
    {
        for (int i = 1; i <= m; ++i)
        {
            cols[i].clear();
        }
        int val;
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &val);
            cols[val].push_back(i);
        }
        int ans = 0;
        for (int i = 1; i <= m; ++i)
        {
            int cnt = cols[i].size();
            int s = 0;
            int e = s;
            int num = 0;
            while (1)
            {
                while (e < cnt && num <= k)
                {
                    ++e;
                    if (e >= cnt)
                    {
                        break;
                    }
                    num += cols[i][e] - cols[i][e - 1] - 1;
                }
                ans = max(ans, e - s);
                if (e >= cnt)
                {
                    break;
                }
                ++s;
                num -= cols[i][s] - cols[i][s - 1] - 1;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}