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 long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int sz = s.size();
    int ans = INT_MAX;
    int ct[10], a[sz], start[10];
    for (int i = 0; i < 10; i++) ct[i] = 0;
    for (int i = 0; i < sz; i++) {
      a[i] = s[i] - '0';
      if (ct[a[i]] == 0) start[a[i]] = i;
      ct[a[i]]++;
    }
    for (int i = 0; i < 10; i++) {
      ans = min(ans, sz - ct[i]);
    }
    for (int i = 0; i < 9; i++) {
      for (int j = i + 1; j < 10; j++) {
        if (ct[i] == 0 or ct[j] == 0) continue;
        int res = 1, last;
        if (start[i] < start[j])
          last = i;
        else
          last = j;
        for (int idx = min(start[i], start[j]) + 1; idx < sz; idx++) {
          if (a[idx] != i and a[idx] != j) continue;
          if (a[idx] != last) {
            res++;
            last = a[idx];
          }
        }
        if (res & 1) res--;
        ans = min(ans, sz - res);
      }
    }
    printf("%d\n", ans);
  }
}
