#include <bits/stdc++.h>
using namespace std;
int solve(const string& s, int x, int y) {
  int res = 0;
  for (auto c : s)
    if (c - '0' == x) {
      ++res;
      swap(x, y);
    }
  if (x != y && res % 2 == 1) --res;
  return res;
}
void solve() {
  string s;
  cin >> s;
  int ans = 0;
  for (int x = 0; x < int(10); ++x)
    for (int y = 0; y < int(10); ++y) ans = max(ans, solve(s, x, y));
  cout << int((s).size()) - ans << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
int solve(const string& s, int x, int y) {
  int res = 0;
  for (auto c : s)
    if (c - '0' == x) {
      ++res;
      swap(x, y);
    }
  if (x != y && res % 2 == 1) --res;
  return res;
}
void solve() {
  string s;
  cin >> s;
  int ans = 0;
  for (int x = 0; x < int(10); ++x)
    for (int y = 0; y < int(10); ++y) ans = max(ans, solve(s, x, y));
  cout << int((s).size()) - ans << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}

```