/*

     题意：一只兔子顺时针跳，另一只逆时针跳，跳石头权值相等而且不能越过起点

     LPS：这道就是LPS的应用，把环倍增成链，套一下LPS，然而并不能理解dp[i][i+n-2] + 1，看别人的解题报告吧，以后来补(玩游戏)

 　　详细解释 */

 /************************************************

 * Author        :Running_Time

 * Created Time  :2015-8-8 16:57:23

 * File Name     :HDOJ_4747_LPS.cpp

  ************************************************/

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <sstream>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <vector>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <list>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cstdlib>

 #include <ctime>

 using namespace std;

 #define lson l, mid, rt << 1

 #define rson mid + 1, r, rt << 1 | 1

 typedef long long ll;

 const int MAXN = 2e3 + ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 int a[MAXN], dp[MAXN][MAXN];

 int main(void)    {     //HDOJ 4745 Two Rabbits

     int n;

     while (scanf ("%d", &n) == )   {

         if (!n) break;

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

             scanf ("%d", &a[i]);

             a[i+n] = a[i];

         }

         memset (dp, , sizeof (dp));

         for (int i=; i<=*n; ++i)  dp[i][i] = ;

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

             for (int j=; j+i-<=*n; ++j)    {

                 if (a[j] == a[j+i-]) dp[j][j+i-] = dp[j+][j+i-] + ;

                 else    {

                     dp[j][j+i-] = max (dp[j+][j+i-], dp[j][j+i-]);

                 }

             }

         }

         int ans = ;

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

             ans = max (ans, dp[i][i+n-]);

             ans = max (ans, dp[i][i+n-] + );

         }

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

     }

     return ;

 }