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 ok[100008];
char a[100008];
int t;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%s", a + 1);
    int len = strlen(a + 1);
    int maxx = 0, ans = len;
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        int num = 0, sum = 0;
        for (int k = 1; k <= len; k++) {
          if (num % 2 == 0) {
            if (a[k] - '0' != i)
              sum++;
            else
              num++;
          } else {
            if (a[k] - '0' != j)
              sum++;
            else
              num++;
          }
        }
        if (i != j && num % 2 == 1) sum++;
        ans = min(ans, sum);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
