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;
int read() {
  char x = getchar();
  int ans = 0, flag = 1;
  while (!isdigit(x))
    if (x == '-')
      flag = -1, x = getchar();
    else
      x = getchar();
  while (isdigit(x)) ans = ans * 10 + x - '0', x = getchar();
  return ans * flag;
}
char s[200005];
signed main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> s;
    int len = strlen(s), ans = 0;
    for (int i = 0; i <= 9; i++) {
      int tmp = 0;
      for (int j = 0; j <= 9; j++) {
        int opt = 0;
        tmp = 0;
        for (int k = 0; k < len; k++) {
          if (!opt) {
            if (s[k] == i + '0') tmp++, opt ^= 1;
          } else {
            if (s[k] == j + '0') tmp++, opt ^= 1;
          }
        }
        if (!opt || i == j) ans = max(ans, tmp);
      }
    }
    cout << len - ans << '\n';
  }
  return 0;
}
