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;
long long gcd(long long a, long long b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int a = 0, b, c, d, m, n, i, j, k, l, x = 0, y = 0, ans = 100000000,
        tot = 0, fl = 0;
    int arr[10];
    memset(arr, 0, sizeof(arr));
    string s;
    cin >> s;
    n = s.size();
    for (i = 0; i < 10; i++) {
      for (j = 0; j < 10; j++) {
        tot = 0;
        fl = i;
        for (k = 0; k < n; k++) {
          if (s[k] - 48 == fl) {
            if (fl == i) {
              fl = j;
            } else if (fl == j)
              fl = i;
          } else
            tot++;
        }
        if (fl == i) ans = min(ans, tot);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
