CodeForces 180E Cubes（vector+尺取）

题目链接：点击打开链接

E. Cubes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's imagine that you're playing the following simple computer game. The screen displaysn lined-up cubes. Each cube is painted one ofm colors. You are allowed to delete not more thank 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.

Examples

Input

10 3 21 2 1 1 3 2 1 1 2 2

Output

4

Input

10 2 21 2 1 2 1 1 2 1 1 2

Output

5

Input

3 1 21 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.

题目大意：n个立方体并列放着，然后有m种颜色，你可以取出0～k个立方体，然后再并拢，问最大的连续相同颜色的立方体有多少个

思路：vector存每个颜色的位置，然后之后用尺取法，最后取最大的就好了

#include <bits/stdc++.h>using namespace std;vector<int>v[111111];int main(){    int n,m,k;    while (~scanf("%d%d%d",&n,&m,&k))    {        for (int i = 1 ; i <= n ; i++ )        {            int a;            scanf("%d",&a);            v[a].push_back(i);        }        int res = 1;        for (int i = 1 ; i <= m ; i++ )        {            int x = v[i].size();            int temp = 1;            int cou = 0;            int y = 0;            for(int j = 1 ; j < x  ; j ++ )            {                temp++;                cou += v[i][j] - v[i][j-1] - 1;                while ( cou > k)                {                    temp--;                    cou -=v[i][y+1] - v[i][y] - 1;                    y++;                }                if ( res < temp)                {                    res = temp;                }                            }        }        printf("%d\n",res);    }}