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 cd = complex<double>;
const double pi = acos(-1);
vector<long long> ft(1, 1LL);
void solve() {
  string s;
  long long n, ret;
  cin >> s;
  n = s.size();
  ret = s.size() - 2;
  for (long long j = 0; j < 10; j++) {
    long long cunt = 0;
    for (auto i : s)
      if (i - '0' == j) cunt++;
    ret = min(ret, n - cunt);
  }
  long long f = 1, f2 = 1;
  for (long long i = 0; i < 10; i++) {
    for (long long j = 0; j < 10; j++) {
      long long cunt = 0;
      f = 1;
      if (i == j) continue;
      for (auto c : s) {
        if (f && c - '0' == i)
          cunt++, f = 1 - f;
        else if (!f && c - '0' == j)
          cunt++, f = 1 - f;
      }
      if (cunt % 2 != 0) cunt = 0;
      ret = min(ret, n - cunt);
    }
  }
  for (long long i = 0; i < s.size(); i++)
    if (s[i] != s[(i + 2) % n]) f2 = 0;
  if (f2) ret = 0;
  cout << ret << '\n';
}
signed main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
