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;
using LL = long long;
constexpr int N = 1e5 + 5;
string s;
int n;
int f(int x, int y) {
  int z = x;
  int res = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == z) {
      res++;
      z = x + y - z;
    }
  }
  if (res & 1) res -= (x != y);
  return res;
}
void solve() {
  cin >> s;
  n = s.size();
  for (char &c : s) c -= '0';
  int res = 0;
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      res = max(res, f(i, j));
    }
  }
  cout << n - res << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
