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 has, arr[20], flag[20][20];
string s;
void f(int a, int b) {
  char p[2], q, now;
  int hit = 0;
  p[0] = char(a + '0'), p[1] = char(b + '0'), now = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == p[now]) {
      if (now == 1) hit += 2;
      now = 1 - now;
    }
  }
  int l = s.size();
  has = min(has, l - hit);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> s;
    has = 1000000009;
    for (int i = 0; i <= 9; i++) arr[i] = 0;
    for (int i = 0; i < s.size(); i++) arr[s[i] - '0']++;
    int k = 0;
    for (int i = 0; i <= 9; i++) k = max(k, arr[i]);
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        if (i == j) continue;
        f(i, j);
        f(j, i);
      }
    }
    int n = s.size();
    has = min(has, min(n - 2, n - k));
    cout << has << endl;
  }
}
