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;
const long long N = 2e5 + 5;
const long long inf = 1e14;
void solve() {
  string s;
  cin >> s;
  long long ans = 1e9;
  for (char c1 = '0'; c1 <= '9'; c1++) {
    for (char c2 = '0'; c2 <= '9'; c2++) {
      long long x = 0, e = 1;
      for (long long i = 0; i < s.length(); i++) {
        if (e) {
          if (s[i] == c1) {
            e = 0;
          } else {
            x++;
          }
        } else {
          if (s[i] == c2) {
            e = 1;
          } else {
            x++;
          }
        }
      }
      if (((long long)s.length() - x) & 1ll) {
        if (c1 == c2) ans = min(ans, x);
      } else
        ans = min(ans, x);
    }
  }
  cout << ans << endl;
}
int32_t main() {
  long long t = 1;
  cin >> t;
  while (t--) solve();
}
