#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  string s;
  cin >> s;
  vector<int> cnt(300);
  for (auto c : s) cnt[c]++;
  int n = s.size();
  int ans = -1;
  for (char c = '0'; c <= '9'; c++) ans = max(ans, cnt[c]);
  ans = n - ans;
  for (char c1 = '0'; c1 <= '9'; c1++) {
    for (char c2 = '0'; c2 <= '9'; c2++) {
      int cur = 0;
      for (auto c : s) {
        if (c == ((cur & 1) ? c1 : c2)) cur++;
      }
      if (cur % 2) cur--;
      ans = min(ans, n - cur);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  string s;
  cin >> s;
  vector<int> cnt(300);
  for (auto c : s) cnt[c]++;
  int n = s.size();
  int ans = -1;
  for (char c = '0'; c <= '9'; c++) ans = max(ans, cnt[c]);
  ans = n - ans;
  for (char c1 = '0'; c1 <= '9'; c1++) {
    for (char c2 = '0'; c2 <= '9'; c2++) {
      int cur = 0;
      for (auto c : s) {
        if (c == ((cur & 1) ? c1 : c2)) cur++;
      }
      if (cur % 2) cur--;
      ans = min(ans, n - cur);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```