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;
int32_t mod = 1e9 + 7;
const long long N = 200005;
const long long INF = (1LL << 60) - 1;
const long long maxn = 200007;
void solveCase() {
  string s;
  cin >> s;
  string a = "0123456789";
  long long ans = 0;
  for (int k1 = 0; k1 < 10; k1++) {
    for (int k2 = 0; k2 < 10; k2++) {
      char x1 = a[k1];
      char x2 = a[k2];
      long long c = 0;
      for (int i = 0; i < s.length(); i++) {
        if (s[i] == x1) {
          c++;
          swap(x1, x2);
        }
      }
      if (x1 != x2 && c % 2) {
        c--;
      }
      ans = max(ans, c);
    }
  }
  cout << s.length() - ans << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solveCase();
}
