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;
const int maxn = 2e5 + 10;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    string s;
    int num[15];
    bool nu[15];
    memset(num, 0, sizeof(num));
    cin >> s;
    int len = s.length();
    for (int i = 0; i < len; i++) {
      num[s[i] - '0']++;
    }
    int maxx = -1;
    int ans = -1;
    for (int i = 0; i < 10; i++) {
      maxx = max(maxx, num[i]);
    }
    if (maxx == 1) {
      cout << len - 2 << endl;
    } else {
      memset(nu, false, sizeof(nu));
      for (int i = 0; i < 10; i++)
        if (num[i] * 2 >= maxx) nu[i] = true;
      for (int i = 0; i < 10; i++) {
        if (!nu[i]) continue;
        for (int j = 0; j < 10; j++) {
          if (!nu[j] || j == i) continue;
          bool f = false;
          int cnt = 0;
          for (int k = 0; k < len; k++) {
            if (s[k] == '0' + j && f) {
              f = false;
              cnt++;
            } else if (s[k] == '0' + i) {
              f = true;
            }
          }
          ans = max(ans, cnt);
        }
      }
      cout << min(len - ans * 2, len - maxx) << endl;
    }
  }
}
