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 oo = (1ll << 60);
int gi() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  cin >> T;
  for (int cas = 1; cas <= T; cas++) {
    string s;
    cin >> s;
    int mp[10] = {0}, len = s.size(), i, j, k, cpl[100] = {0};
    vector<vector<int> > v(10);
    int mxsingle = -1, mxcpl = -1;
    for (i = 0; i < len; i++) {
      mp[s[i] - '0']++;
      if (mp[s[i] - '0'] > mxsingle) mxsingle = mp[s[i] - '0'];
      v[s[i] - '0'].push_back(i);
    }
    for (i = 0; i < 10; i++) v[i].push_back(len);
    for (i = 0; i < 10; i++) {
      for (j = 0; j < v[i].size() - 1; j++) {
        for (k = 0; k < 10; k++) {
          if (k == i) continue;
          int id =
              lower_bound(v[k].begin(), v[k].end(), v[i][j]) - v[k].begin();
          if (id == v[k].size() || id == v[k].size() - 1) continue;
          if (v[k][id] > v[i][j + 1]) continue;
          cpl[i * 10 + k]++;
        }
      }
    }
    for (i = 0; i < 100; i++)
      if (mxcpl < cpl[i]) mxcpl = cpl[i];
    int mx = max(mxsingle, 2 * mxcpl);
    cout << len - mx << '\n';
  }
  return 0;
}
