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 = 1e9 + 112;
int t, cnt[11], n;
string s;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> s;
    n = s.size();
    memset(cnt, 0, sizeof(cnt));
    vector<int> pos[10];
    for (int i = (int)0; i < (int)n; ++i) {
      ++cnt[s[i] - '0'];
      pos[s[i] - '0'].push_back(i);
    }
    int ans = inf;
    for (int i = (int)0; i <= (int)9; ++i) ans = min(ans, n - cnt[i]);
    for (int i = (int)0; i <= (int)9; ++i)
      for (int j = (int)0; j <= (int)9; ++j) {
        if (i == j) continue;
        int p = 0, t = 0, bef = -1;
        for (int k = (int)0; k < (int)pos[i].size(); ++k) {
          if (pos[i][k] < bef) continue;
          for (int l = (int)p; l <= (int)pos[j].size() - 1; ++l) {
            if (pos[j][l] > pos[i][k]) {
              ++t;
              p = l + 1;
              bef = pos[j][l];
              break;
            }
          }
        }
        ans = min(ans, n - 2 * t);
      }
    cout << ans << '\n';
  }
}
