#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long mod = 1e9 + 7;
const int N = 1e5 + 5;
int T, n, m, k, q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    n = s.size();
    int mx = 0;
    for (int x = 0; x < 10; x++) {
      for (int y = 0; y < 10; y++) {
        int mark = 0;
        int cnt = 0;
        for (int i = 0; i < n; i++) {
          if (s[i] - '0' == x && mark == 0)
            cnt++, mark = 1;
          else if (s[i] - '0' == y && mark == 1)
            cnt++, mark = 0;
        }
        if (cnt % 2 == 1 && x != y) cnt--;
        mx = max(mx, cnt);
      }
    }
    cout << n - mx << "\n";
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int inf = 1e9 + 7;
const long long mod = 1e9 + 7;
const int N = 1e5 + 5;
int T, n, m, k, q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    n = s.size();
    int mx = 0;
    for (int x = 0; x < 10; x++) {
      for (int y = 0; y < 10; y++) {
        int mark = 0;
        int cnt = 0;
        for (int i = 0; i < n; i++) {
          if (s[i] - '0' == x && mark == 0)
            cnt++, mark = 1;
          else if (s[i] - '0' == y && mark == 1)
            cnt++, mark = 0;
        }
        if (cnt % 2 == 1 && x != y) cnt--;
        mx = max(mx, cnt);
      }
    }
    cout << n - mx << "\n";
  }
}

```