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;
const int inf = 0x3f3f3f3f;
const long long inff = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 1e5 + 5;
long long a[10005];
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int cou[10];
    for (int i = 0; i < 10; i++) cou[i] = 0;
    for (int i = 0; i < s.size(); ++i) cou[s[i] - '0']++;
    int ma = 0;
    for (int i = 0; i <= 9; i++) {
      if (cou[i] >= ma) {
        ma = cou[i];
      }
    }
    if (ma == 1) {
      cout << s.size() - 2 << endl;
      continue;
    }
    int maxx = 0;
    for (int x = 0; x <= 9; x++) {
      for (int y = 0; y <= 9; y++) {
        int f = 1, sum = 0;
        ;
        for (int i = 0; i < s.size(); i++) {
          int temp = (f == 1 ? x : y);
          if (s[i] == '0' + temp) {
            sum++;
            f = -f;
          }
        }
        if (sum % 2 == 0 || x == y)
          maxx = max(maxx, sum);
        else
          maxx = max(maxx, sum - 1);
      }
    }
    cout << s.size() - maxx << endl;
  }
  return 0;
}
