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 INF = 1e9 + 5;
void self_max(long long& a, const long long& b) { a = max(a, b); }
void self_min(long long& a, const long long& b) { a = min(a, b); }
inline void preprocessing() {}
long long getMaxAns(const string& s, long long x, long long y) {
  long long ans = 0;
  for (long long i = 0; i < s.size(); ++i) {
    if (s[i] - '0' == x) {
      ++ans;
      swap(x, y);
    }
  }
  if (x != y && ans % 2 != 0) {
    --ans;
  }
  return ans;
}
inline void solve() {
  string s;
  cin >> s;
  long long ans = 0;
  for (long long i = 0; i < 10; ++i)
    for (long long j = 0; j < 10; ++j) self_max(ans, getMaxAns(s, i, j));
  cout << s.size() - ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  preprocessing();
  long long t;
  cin >> t;
  while (t--) solve();
}
