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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    string S;
    cin >> S;
    int i = 0, n = S.size();
    int A[10] = {0}, C[10] = {0}, sum = 0, sum1 = 0;
    vector<bool> B(10);
    while (S[i] != '\0') {
      A[S[i] - '0']++;
      i++;
    }
    for (int i = 0; i < 10; i++) {
      sum = max(sum, A[i]);
    }
    for (int i = 0; i < 10; i++) {
      sum1 = 0;
      for (int j = 0; j < n;) {
        for (int k = 0; k < 10; k++) B[k] = 0;
        if (S[j] - '0' == i) {
          j++;
          while (S[j] - '0' != i && j < n) {
            if (!B[S[j] - '0']) {
              C[S[j] - '0'] += 2;
              B[S[j] - '0'] = 1;
            }
            j++;
          }
        } else
          j++;
      }
      for (int j = 0; j < 10; j++) {
        sum = max(sum, C[j]);
        C[j] = 0;
      }
    }
    cout << n - sum << "\n";
  }
  return 0;
}
