//#pragma comment(linker, "/STACK:102400000, 102400000")

 #include <algorithm>

 #include <iostream>

 #include <cstdlib>

 #include <cstring>

 #include <cstdio>

 #include <vector>

 #include <cmath>

 #include <ctime>

 #include <list>

 #include <set>

 #include <map>

 using namespace std;

 typedef long long LL;

 typedef pair <int, int> P;

 const int N = 1e5 + ;

 vector <int> ans[N];

 int a[N*];

 int solve(int u, int k) {

     int len = ans[u].size(), notcnt = , cnt = , l = , res = ;

     //notcnt表示中间删除数的多少，cnt表示颜色为u的序列的长度

     for(int i = ; i < len; ++i) {

         int v = ans[u][i];

         notcnt += v - ans[u][i - ] - ;

         cnt++;

         while(notcnt > k && l <= i) {

             notcnt -= ans[u][l + ] - ans[u][l] - ;

             cnt--;

             l++;

         }

         res = max(cnt, res);

     }

     return res;

 }

 int main()

 {

     int n, m, k;

     scanf("%d %d %d", &n, &m, &k);

     for(int i = ; i <= n; ++i) {

         scanf("%d", a + i);

         ans[a[i]].push_back(i);

     }

     int res = ;

     for(int i = ; i <= m; ++i) {

         if(ans[i].size()) { //要是存在颜色为i的数

             res = max(res, solve(i, k));

         }

     }

     printf("%d\n", res);

     return ;

 }