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;
double mod(double a) {
  if (a < 0) return -1 * a;
  return a;
}
bool prime[100000000 + 1];
void SieveOfEratosthenes() {
  long long int n = 100000000;
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1, u, j;
  cin >> t;
  for (u = 1; u <= t; u++) {
    long long int m, n, i, j, a, b, x, y, k, p = 0, q = 0, p1 = INT_MAX, p2 = 0,
                                             p3 = 0, ans = 0, l, r, index = -1;
    string s;
    cin >> s;
    for (i = 0; i <= 9; i++) {
      for (j = 0; j <= 9; j++) {
        char c1 = i + '0';
        char c2 = j + '0';
        p = 0;
        for (k = 0; k < s.size(); k++) {
          if (p % 2 == 0) {
            if (s[k] == c1) p++;
          } else {
            if (s[k] == c2) p++;
          }
        }
        if (i != j && p % 2 != 0) p--;
        p2 = max(p, p2);
      }
    }
    cout << s.size() - p2 << endl;
  }
}
