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;
const int N = 200005;
int mod = 1e9 + 7;
string s;
int n, t;
int calc(int x, int y) {
  int val = 0;
  for (int i = 0; i < s.size(); i++) {
    int digit = s[i] - '0';
    if (digit == x) val++, swap(x, y);
  }
  return val;
}
int even() {
  int val = 0;
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      int temp = calc(i, j);
      if (temp % 2 == 0) val = max(val, temp);
      if (temp % 2 && i == j) val = max(val, temp);
    }
  }
  return val;
}
int main() {
  cin >> t;
  int ans;
  while (t--) {
    cin >> s;
    cout << s.size() - even() << '\n';
  }
  return 0;
}
