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;
string st;
long long getAns(long long a, long long b) {
  long long preOk = 0, ans = 0, i;
  if (a == b) {
    for (i = 0; i < st.size(); i++) ans += st[i] == a;
  } else {
    for (i = 0; i < st.size(); i++) {
      if (preOk && st[i] == b) ans += 2, preOk = 0;
      if (st[i] == a) preOk = 1;
    }
  }
  return ans;
}
void solve() {
  long long n, a, b, i, j, mx = 0;
  cin >> st;
  vector<long long> v;
  for (i = '0'; i <= '9'; i++) {
    for (j = '0'; j <= '9'; j++) {
      mx = max(mx, getAns(i, j));
    }
  }
  long long ans = st.size() - mx;
  cout << ans << "\n";
}
int main() {
  long long t, i;
  cin >> t;
  while (t--) solve();
  return 0;
}
