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 get_temp(string& str, char x, char y, int n) {
  int cnt = 0;
  int i = 0;
  char main = x;
  while (i < n) {
    if (main == str[i]) {
      cnt++;
      main = (main == x) ? y : x;
      i++;
      continue;
    }
    i++;
  }
  if (cnt % 2 == 1) {
    cnt--;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    int len = str.length();
    int arr[15];
    for (int i = 0; i <= 10; i++) {
      arr[i] = 0;
    }
    for (int i = 0; i < len; i++) {
      arr[str[i] - '0']++;
    }
    int maxi = 0;
    for (int i = 0; i <= 9; i++) {
      maxi = max(arr[i], maxi);
    }
    int ans = maxi;
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        if (i == j) {
          continue;
        }
        int temp = get_temp(str, i + '0', j + '0', len);
        maxi = max(maxi, temp);
      }
    }
    cout << len - maxi << "\n";
  }
  return 0;
}
