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 t;
char s[200005];
int tong[205];
int len;
char ch1, ch2;
int find1;
int max1 = 1000000;
int main() {
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> s;
    len = strlen(s);
    for (int j = 0; j < len; j++) {
      tong[s[j] - '0']++;
    }
    for (int k = 1; k <= 99; k++) {
      ch1 = k / 10 + '0';
      ch2 = k % 10 + '0';
      if (ch1 == ch2) continue;
      for (int j = 0; j < len; j++) {
        if (s[j] == ch1 && !find1) {
          find1 = 1;
        }
        if (s[j] == ch2 && find1) {
          tong[k + 100]++;
          find1 = 0;
        }
      }
      find1 = 0;
    }
    for (int i = 0; i < 200; i++) {
      if (i < 10) {
        max1 = min(max1, len - tong[i]);
      } else {
        max1 = min(max1, len - 2 * tong[i]);
      }
    }
    cout << max1 << endl;
    max1 = 10000000;
    memset(tong, 0, sizeof(tong));
  }
}
