#include <bits/stdc++.h>
using namespace std;
int f(char s) { return (int)(s - '0'); }
int solve() {
  string s;
  cin >> s;
  int a[10][10], count[10][10];
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      a[i][j] = 0;
      count[i][j] = 0;
    }
  }
  for (int i = 0; i < int((s).size()); i++) {
    for (int j = 0; j <= 9; j++) {
      if (f(s[i]) == j) {
        if (count[f(s[i])][j] == 0) {
          count[f(s[i])][j] = 1;
          a[j][j]++;
        } else {
          count[f(s[i])][j] = 0;
          a[j][j]++;
        }
        continue;
      }
      if (count[f(s[i])][j] == 0) {
        count[f(s[i])][j] = 1;
      }
      if (count[j][f(s[i])] == 1) {
        count[j][f(s[i])] = 0;
        a[j][f(s[i])] += 2;
      }
    }
  }
  int mx = 0;
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      if (a[i][j] > mx) mx = a[i][j];
    }
  }
  cout << int((s).size()) - mx << "\n";
  return 0;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int f(char s) { return (int)(s - '0'); }
int solve() {
  string s;
  cin >> s;
  int a[10][10], count[10][10];
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      a[i][j] = 0;
      count[i][j] = 0;
    }
  }
  for (int i = 0; i < int((s).size()); i++) {
    for (int j = 0; j <= 9; j++) {
      if (f(s[i]) == j) {
        if (count[f(s[i])][j] == 0) {
          count[f(s[i])][j] = 1;
          a[j][j]++;
        } else {
          count[f(s[i])][j] = 0;
          a[j][j]++;
        }
        continue;
      }
      if (count[f(s[i])][j] == 0) {
        count[f(s[i])][j] = 1;
      }
      if (count[j][f(s[i])] == 1) {
        count[j][f(s[i])] = 0;
        a[j][f(s[i])] += 2;
      }
    }
  }
  int mx = 0;
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      if (a[i][j] > mx) mx = a[i][j];
    }
  }
  cout << int((s).size()) - mx << "\n";
  return 0;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```