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;
long long md(string& s, long long x, long long y) {
  long long rv = 0;
  bool oy = false;
  for (long long i = 0; i < (long long)s.size(); ++i) {
    if (oy) {
      if (s[i] == '0' + y) {
        oy = !oy;
      } else {
        rv++;
      }
    } else {
      if (s[i] == '0' + x) {
        oy = !oy;
      } else {
        rv++;
      }
    }
  }
  rv += oy;
  return rv;
}
long long C(string& s) {
  vector<long long> c(10, 0);
  for (auto si : s) {
    c[si - '0']++;
  }
  long long rv = s.size() - 2;
  for (long long i = 0; i <= 9; ++i) {
    rv = min(rv, ((long long)s.size()) - c[i]);
    for (long long j = 0; j < i; ++j) {
      rv = min(rv, md(s, i, j));
    }
    for (long long j = i + 1; j <= 9; ++j) {
      rv = min(rv, md(s, i, j));
    }
  }
  return rv;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  for (; t-- > 0;) {
    string s;
    cin >> s;
    cout << C(s) << endl;
  }
  return 0;
}
