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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string x;
    cin >> x;
    int cnt[10] = {0};
    for (int i = 0; i < x.length(); ++i) cnt[x[i] - '0']++;
    vector<int> temp;
    int sav = 0;
    for (int i = 0; i <= 9; ++i) {
      if (cnt[i]) temp.push_back(i);
      if ((cnt[i] & 1) && cnt[i] > sav) sav = cnt[i];
    }
    int ans = x.length() - sav;
    for (int i = 0; i < temp.size(); ++i)
      for (int j = 0; j < temp.size(); ++j) {
        int ret = 0, now = 0;
        for (int k = 0; k < x.length(); ++k) {
          if (now & 1) {
            if (x[k] != temp[j] + '0')
              ++ret;
            else
              ++now;
          } else {
            if (x[k] != temp[i] + '0')
              ++ret;
            else
              ++now;
          }
        }
        ans = min(ans, ret + (now & 1));
      }
    cout << ans << endl;
  }
  return 0;
}
