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;
long long powmod(long long a, long long b) {
  long long res = 1;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = (int)(s).size();
    int ans = 0;
    vector<int> cnt[10];
    for (int i = 0; i < n; ++i) cnt[s[i] - 48].push_back(i);
    for (int i = 0; i < 10; ++i) ans = max(ans, (int)(cnt[i]).size());
    for (int i = 0; i < 10; ++i) {
      for (int j = 0; j < 10; ++j) {
        if (i == j) continue;
        if ((int)(cnt[i]).size() == 0 || (int)(cnt[j]).size() == 0) continue;
        int a = 0, b = 0, ni = 1, prev = -1, t = 0;
        while ((ni && a < (int)(cnt[i]).size()) ||
               (ni == 0 && b < (int)(cnt[j]).size())) {
          while (ni && a < (int)(cnt[i]).size() && cnt[i][a] < prev) ++a;
          if (ni && a < (int)(cnt[i]).size()) {
            ni = 0;
            prev = cnt[i][a++];
            t++;
            continue;
          }
          while (ni == 0 && b < (int)(cnt[j]).size() && cnt[j][b] < prev) ++b;
          if (!ni && b < (int)(cnt[j]).size()) {
            ni = 1;
            prev = cnt[j][b++];
            t++;
          }
        }
        if ((t & 1) == 0) ans = max(ans, t);
      }
    }
    ans = n - ans;
    cout << ans << '\n';
  }
  return 0;
}
