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 N = 300;
void print() { cout << "\n"; }
template <typename T, typename... Types>
void print(const T& arg, const Types&... args) {
  cout << arg << " ";
  print(args...);
}
long long power(long long a, long long b) {
  if (!b) return 1;
  return (power((a * a), b >> 1) * ((b & 1) ? a : 1));
}
long long power(long long a, long long b, long long p) {
  if (!b) return 1;
  return (power((a * a) % p, b >> 1, p) * ((b & 1) ? a : 1)) % p;
}
long long solve(string s, long long a, long long b) {
  long long r = 0;
  for (auto ch : s) {
    if (ch - '0' == a) {
      r++;
      swap(a, b);
    }
  }
  if (a != b && r & 1) r--;
  return r;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long result = 0;
    for (__typeof(10) i = (0) - ((0) > (10)); i != (10) - ((0) > (10));
         i += 1 - 2 * ((0) > (10))) {
      for (__typeof(10) j = (0) - ((0) > (10)); j != (10) - ((0) > (10));
           j += 1 - 2 * ((0) > (10)))
        result = max(result, solve(s, i, j));
    }
    print(s.length() - result);
  }
}
