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;
long long int po(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return po((x * x) % 1000000007, n / 2);
  else
    return (x * po((x * x) % 1000000007, (n - 1) / 2)) % 1000000007;
}
long long int GCD(long long int A, long long int B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int f[10] = {0};
    long long int n = s.length();
    long long int an = n - 1;
    for (int i = 0; i < n; ++i) f[s[i] - '0']++;
    for (int i = 0; i < 10; ++i) an = min(an, n - f[i]);
    for (int i = 0; i < 10; ++i) {
      for (int j = 0; j < 10; ++j) {
        if (i == j) continue;
        long long int d = 0;
        long long int ct = 0;
        for (int k = 0; k < n; ++k) {
          if (s[k] - '0' == i && d % 2 == 0) {
            ct++;
            d++;
          }
          if (s[k] - '0' == j && d % 2 == 1) {
            ct++;
            d++;
          }
        }
        ct -= ct % 2;
        an = min(an, n - ct);
      }
    }
    cout << an << "\n";
  }
}
