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;
template <class T>
void print(vector<T>& v) {
  for (T x : v) cout << x << " ";
  cout << "\n";
}
void inp(vector<int>& v, int n) {
  for (int i = 0; i < n; ++i) scanf("%d ", &v[i]);
}
void sol() {
  string s;
  cin >> s;
  int n = s.size();
  int fr[10] = {0};
  for (char c : s) {
    fr[c - '0']++;
  }
  int mx = 0, cnt, x;
  for (int i : fr) mx = max(i, mx);
  int ans = n - mx;
  for (char i = '0'; i <= '9'; i++) {
    for (char j = '0'; j <= '9'; j++) {
      if (i == j) continue;
      cnt = 0;
      for (char c : s) {
        if (c == i && (cnt & 1) == 0)
          cnt++;
        else if (c == j && (cnt & 1))
          cnt++;
      }
      if (cnt & 1) cnt--;
      ans = min(ans, n - cnt);
    }
  }
  printf("%d\n", ans);
  return;
}
int main() {
  int t;
  cin >> t;
  t++;
  for (int i = 1; i < t; ++i) {
    sol();
  }
  return 0;
}
