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;
void solve() {
  long long int n, i, j, k, l, m, o, p[10], q, r, s, t;
  string a;
  memset(p, 0, sizeof(p));
  cin >> a;
  n = a.length();
  long long int ans = LLONG_MAX;
  for (i = 0; i < n; i++) {
    m = a[i];
    m -= 48;
    p[m]++;
    ans = min(ans, n - p[m]);
  }
  for (i = 0; i <= 9; i++) {
    for (j = 0; j <= 9; j++) {
      l = 0;
      string ss = "";
      for (k = 0; k < n; k++) {
        if (a[k] == i + '0' && ss.length() == 0) {
          ss += a[k];
        } else if (a[k] == i + '0' && ss.length() > 0 &&
                   ss[ss.length() - 1] == j + '0') {
          ss += a[k];
        } else if (a[k] == j + '0' && ss.length() > 0 &&
                   ss[ss.length() - 1] == i + '0') {
          ss += a[k];
        }
      }
      o = ss.length();
      if (o % 2) o--;
      ans = min(ans, n - o);
    }
  }
  cout << ans << endl;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
