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 mod = 998244353;
const long long N = 1000000007;
long long binpow(long long b, long long p) {
  long long ans = 1;
  while (p > 0) {
    if (p & 1) ans = (ans * b) % mod;
    b = (b * b) % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long x) { return binpow(x, mod - 2); }
long long divide(long long x, long long y) {
  return ((x % mod) * (inv(y))) % mod;
}
long long solve(string& s, char c, char d) {
  int cur = 0;
  long long ans = 0;
  long long i;
  for (i = 0; i < s.size(); i++) {
    if (cur == 0) {
      if (s[i] != c)
        ans++;
      else
        cur ^= 1;
    } else {
      if (s[i] != d)
        ans++;
      else
        cur ^= 1;
    }
  }
  long long check = s.size() - ans;
  if (check % 2 != 0 && (c != d)) {
    return ans + 1;
  } else
    return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tt, i, n;
  cin >> tt;
  while (tt--) {
    string s;
    cin >> s;
    long long ans, val1, val2;
    if (s.size() <= 2) {
      cout << 0 << endl;
      continue;
    } else
      ans = s.size() - 2;
    char c, d;
    for (c = '0'; c <= '9'; c++) {
      for (d = '0'; d <= '9'; d++) {
        val1 = solve(s, c, d);
        ans = min(val1, ans);
      }
    }
    cout << ans << endl;
  }
  return 0;
}
