/*

     题意：给一个字符串，连续相同的段落可以合并，gogogo->3(go)，问最小表示的长度

     区间DP：dp[i][j]表示[i,j]的区间最小表示长度，那么dp[i][j] = min (dp[j][k] + dp[k+1][i+j-1]),

             digit (i / k) + dp[j][j+k-1] + 2)后者表示可以压缩成k长度连续相同的字符串
 4.5　　详细解释 */

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

 * Author        :Running_Time

 * Created Time  :2015-8-12 15:28:11

 * File Name     :UVA_1351.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 = 2e2 + ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 char str[MAXN];

 int dp[MAXN][MAXN];

 bool check(int l, int r, int k)   {

     int i = ;

     while (i < k)   {

         for (int p=; l+p*k+i<=r; ++p)  {

             if (str[l+i] != str[l+p*k+i])   return  false;

         }

         ++i;

     }

     return true;

 }

 int digit(int x)    {

     int ret = ;

     while (x)   {

         ret++;  x /= ;

     }

     return ret;

 }

 int main(void)    {     //UVA 1351 String Compression

     int T;  scanf ("%d", &T);

     while (T--) {

         scanf ("%s", str + );

         int len = strlen (str + );

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

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

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

                 dp[j][j+i-] = INF;

                 int &x = dp[j][j+i-];

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

                     x = min (x, dp[j][k] + dp[k+][i+j-]);

                 }

                 for (int k=; k<=i/; ++k)  {

                     if (i % k != ) continue;

                     if (check (j, i + j - , k))    {

                         x = min (x, digit (i / k) + dp[j][j+k-] + );

                     }

                 }

             }

         }

         printf ("%d\n", dp[][len]);

     }

     return ;

 }