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;
void INPUT() {}
long long power(long long n, long long p) {
  long long res = 1LL;
  while (p) {
    if (p & 1) res *= n;
    res %= 1000000007;
    n *= n;
    n %= 1000000007;
    p >>= 1LL;
  }
  return res;
}
void goodString() {
  string s;
  cin >> s;
  bool same = true;
  for (long long i = 1; i < s.length(); i++) {
    if (s[i] != s[i - 1]) {
      same = false;
      break;
    }
  }
  if (same) {
    cout << 0 << "\n";
    return;
  }
  long long mx = 0;
  bool fi = false, fj = false;
  long long temp = 0;
  for (long long i = 0; i < 10; i++) {
    for (long long j = 0; j < 10; j++) {
      temp = 0;
      fi = false;
      for (long long k = 0; k < s.length(); k++) {
        if (i == j && s[k] - '0' == i) {
          temp++;
          continue;
        }
        if ((s[k] - '0') == j && fi) {
          temp++;
          fi = false;
        }
        if ((s[k] - '0') == i && !fi) {
          temp++;
          fi = true;
        }
      }
      if (i != j) temp = (temp % 2 == 0) ? temp : temp - 1;
      if (temp >= 0) mx = max(mx, temp);
    }
  }
  cout << s.length() - mx << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  for (int i = 0; i <= t - 1; i++) {
    goodString();
  }
  return 0;
}
