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;
using namespace std::chrono;
long long int powD(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % (long long int)(1e9 + 7);
    x = (x * x) % (long long int)(1e9 + 7);
    y = y >> 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    string second;
    cin >> second;
    int n = second.size();
    vector<vector<int>> idx(10);
    for (int i = 0; i < n; ++i) {
      idx[second[i] - '0'].push_back(i);
    }
    int minm = n - 2;
    for (int i = 0; i < 10; ++i) {
      int sz = idx[i].size();
      minm = min(minm, n - sz);
    }
    for (int i = 0; i < 10; ++i) {
      if (idx[i].size() == 0) continue;
      for (int j = 0; j < 10; ++j) {
        if (i == j || idx[j].size() == 0) continue;
        int cnt = 0;
        for (int k = 0; k < n; ++k) {
          if (cnt % 2 == 0) {
            if (second[k] == '0' + i) cnt++;
          } else {
            if (second[k] == '0' + j) cnt++;
          }
        }
        if (cnt % 2) cnt--;
        minm = min(minm, n - cnt);
      }
    }
    cout << minm << "\n";
  }
}
