Let's call left cyclic shift of some string t_1 t_2 t_3 ... t_{n - 1} t_n as string t_2 t_3 ... t_{n - 1} t_n t_1.

Analogically, let's call right cyclic shift of string t as string t_n t_1 t_2 t_3 ... t_{n - 1}.

Let's say string t is good if its left cyclic shift is equal to its right cyclic shift.

You are given string s which consists of digits 0–9.

What is the minimum number of characters you need to erase from s to make it good?

Input

The first line contains single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Next t lines contains test cases — one per line. The first and only line of each test case contains string s (2 ≤ |s| ≤ 2 ⋅ 10^5). Each character s_i is digit 0–9.

It's guaranteed that the total length of strings doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the minimum number of characters you need to erase from s to make it good.

Example

Input


3
95831
100120013
252525252525


Output


3
5
0

Note

In the first test case, you can erase any 3 characters, for example, the 1-st, the 3-rd, and the 4-th. You'll get string 51 and it is good.

In the second test case, we can erase all characters except 0: the remaining string is 0000 and it's good.

In the third test case, the given string s is already good.
#include <bits/stdc++.h>
using namespace std;
char s[200005];
int dp[200005][11], ss[200005], f[11][200005], tt[11];
int main() {
  int t;
  cin >> t;
  while (t--) {
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    for (int i = 1; i <= n; i++) ss[i] = s[i] - '0';
    for (int i = 0; i < 10; i++) f[i][0] = 0;
    memset(dp[1], 0, sizeof(dp[1]));
    f[ss[1]][++f[ss[1]][0]] = 1;
    for (int i = 2; i <= n; i++) {
      memset(dp[i], 0, sizeof(dp[i]));
      for (int j = 0; j < 10; j++) {
        if (!f[j][0]) {
          dp[i][j] = 0;
          continue;
        }
        int wz = lower_bound(f[ss[i]] + 1, f[ss[i]] + f[ss[i]][0] + 1,
                             f[j][f[j][0]]) -
                 f[ss[i]] - 1;
        if (!wz) {
          dp[i][j] = 2;
          continue;
        }
        dp[i][j] = dp[f[ss[i]][wz]][j] + 2;
      }
      f[ss[i]][++f[ss[i]][0]] = i;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
      for (int j = 0; j < 10; j++) ans = max(ans, dp[i][j]);
    memset(tt, 0, sizeof(tt));
    for (int i = 1; i <= n; i++) tt[ss[i]]++;
    for (int i = 0; i < 10; i++) ans = max(ans, tt[i]);
    cout << n - ans << endl;
  }
  return 0;
}
