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;
const long long MOD = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long t, n, l, r;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    n = s.length();
    long long f[15] = {0};
    long long mx = 0;
    for (auto c : s) {
      f[c - '0']++;
    }
    for (long long i = 0; i < 10; i++) mx = max(mx, f[i]);
    for (long long i = 0; i < 10; i++) {
      string tmp = "";
      long long len = 0;
      char curr = '0' + i;
      char prev;
      for (long long j = 0; j < 10; j++) {
        len = 0;
        if (i == j) continue;
        prev = '0' + j;
        for (char ch : s) {
          if (len & 1) {
            if (ch == prev) len++;
          } else {
            if (ch == curr) len++;
          }
        }
        if (len % 2 == 0) mx = max(mx, len);
      }
    }
    cout << n - mx << "\n";
  }
  return 0;
}
