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 n, k, z;
long long a[200005];
long long pref[200005] = {0};
long long solve(long long i) {
  long long ans = 0;
  long long ck = k;
  if (ck < i) return 0;
  if (ck >= i) {
    ans = pref[min(i, ck)];
    ck = ck - i;
  }
  if (ck > 0) {
    if (2 * z <= ck) {
      ans += z * (a[i] + a[i + 1]);
      ck = ck - 2 * z;
    } else {
      ans += ((ck + 1) / 2) * a[i + 1] + (ck - (ck + 1) / 2) * a[i];
      ck = 0;
    }
  }
  if (ck > 0) {
    ans += pref[ck + i] - pref[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long sze = s.size();
    long long ans = INT_MAX;
    long long curr = 0;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        curr = 0;
        char c = '0' + i;
        for (int k = 0; k < sze; k++) {
          if (s[k] == c) {
            if (c == '0' + i)
              c = '0' + j;
            else
              c = '0' + i;
            curr++;
          }
        }
        if (curr % 2 && i != j) curr--;
        ans = min(ans, sze - curr);
      }
    }
    cout << ans << endl;
  }
}
