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;
int gcd(int a, int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
int main() {
  int ttt;
  cin >> ttt;
  while (ttt--) {
    string s;
    cin >> s;
    int n = s.length();
    if (n == 1 || n == 2) {
      cout << 0 << endl;
      continue;
    }
    int count[10];
    for (int i = 0; i < 10; i++) count[i] = 0;
    for (int i = 0; i < n; i++) count[s[i] - '0']++;
    int ans = n - 2;
    for (int i = 0; i < 10; i++) ans = min(ans, n - count[i]);
    for (int x = 0; x < 10; x++) {
      for (int j = 0; j < 10; j++) {
        int temp = 0, flag = 0;
        for (int i = 0; i < n; i++) {
          if (flag && (s[i] - '0') == j) {
            temp++;
            flag = ~flag;
          } else if (!flag && (s[i] - '0') == x) {
            temp++;
            flag = ~flag;
          }
        }
        ans = min(ans, n - temp + temp % 2);
      }
    }
    cout << ans << endl;
  }
  return 0;
}
