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 = 1000000007;
const long long INF = 0x3f3f3f3f;
const long double PI = 4 * atan((long double)1);
void setIO(string y) {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  freopen((y + ".in").c_str(), "r", stdin);
  freopen((y + ".out").c_str(), "w", stdout);
}
long long second(string &s, long long a, long long b) {
  long long R = 0;
  for (long long i = 0; i < (long long)s.size(); ++i) {
    char c = s[i];
    if (c - 48 == a) {
      ++R;
      swap(a, b);
    }
  }
  if (a != b && R % 2) --R;
  return R;
}
void solve() {
  string s;
  cin >> s;
  long long R = 0;
  for (long long i = 0; i <= 9; ++i)
    for (long long j = 0; j <= 9; ++j) R = max(R, second(s, i, j));
  cout << (long long)s.size() - R << '\n';
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
}
