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 N = 500 * 1000 + 9;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
void HakunaMatata() {
  string s;
  cin >> s;
  long long n = (long long)s.size();
  long long mx = 0;
  for (char a = '0'; a <= '9'; a++) {
    for (char b = '0'; b <= '9'; b++) {
      long long len = 0;
      if (a != b) {
        string t = "";
        t.push_back(a);
        t.push_back(b);
        long long j = 0;
        for (long long i = 0; i < n; i++) {
          if (s[i] == t[j]) {
            if (j == 1) {
              len = len + 2;
              j = 0;
            } else {
              j++;
            }
          }
        }
      } else {
        for (long long i = 0; i < n; i++)
          if (s[i] == a) len++;
      }
      mx = max(mx, len);
    }
  }
  cout << n - mx << '\n';
  return;
}
signed main() {
  long long ___T;
  scanf("%lld", &___T);
  for (long long cs = 1; cs <= ___T; cs++) HakunaMatata();
  return 0;
}
