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 int N = 5000 * 100 + 5;
bool isPrime(int a) {
  for (int i = 2; i * i <= a; ++i)
    if (a % i == 0) return false;
  return true;
}
bool ss(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.second < b.second);
}
long long ss(string s, long long x, long long y) {
  long long res = 0;
  for (auto c : s)
    if (c - '0' == x) {
      res++;
      swap(x, y);
    }
  if (x != y && res % 2 == 1) res--;
  return res;
}
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++) {
      ans = max(ans, ss(s, i, j));
    }
  }
  cout << s.length() - ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  ;
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
