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 MAXN = 1e9;
long double pi = 3.1415926536;
long long mod = 1e9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    long long int n, i;
    map<long long int, long long int> even;
    cin >> s;
    n = s.length();
    for (i = 0; i < n; i++) {
      even[s[i] - 48]++;
    }
    long long int mx = 0;
    for (auto j : even) {
      mx = max(mx, j.second);
    }
    long long int mxe = 0, j;
    for (i = 0; i < 10; i++) {
      long long int m = 0, f = 0;
      for (long long int k = 0; k < 10; k++) {
        m = 0, f = 0;
        for (j = 0; j < n; j++) {
          if (f == 1) {
            if (k == s[j] - 48) {
              m++;
              f = 0;
            }
          } else {
            if (i == s[j] - 48) {
              m++;
              f = 1;
            }
          }
        }
        if (m % 2 == 0) {
          mxe = max(m, mxe);
        }
      }
    }
    cout << min(n - mx, n - mxe) << endl;
  }
}
