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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie();
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long n = s.length();
    map<long long, long long> mp;
    map<long long, long long> mp2;
    long long max1 = 0;
    long long max2 = 0;
    long long val = 0;
    for (long long i = 0; i < n; i++) {
      mp[s[i] - '0']++;
      max1 = max(max1, mp[s[i] - '0']);
      if (max1 < mp[s[i] - '0']) {
        max1 = mp[s[i] - '0'];
        val = s[i] - '0';
      }
    }
    long long ans = 2000001;
    long long warna = n - (max1);
    if (max1 == 1) {
      ans = n - 2;
      if (ans < 0) {
        ans = 0;
      }
    }
    if (n == 1) {
      ans = 0;
    }
    long long ok = min(ans, warna);
    long long count = 0, f = 0, f2 = 0, flag = 0, tt = 0;
    f = 0;
    f2 = 0;
    for (long long one = 0; one < 10; one++) {
      for (long long two = 0; two < 10; two++) {
        if (two != one) {
          f = 0;
          f2 = 0;
          flag = 0;
          for (long long i = 0; i < n; i++) {
            if (flag == 0) {
              if (s[i] - '0' == one) {
                f++;
                flag = flag ^ 1;
              }
            }
            if (flag == 1) {
              if (s[i] - '0' == two) {
                f2++;
                flag = flag ^ 1;
              }
            }
          }
          tt = (f2 + f) / 2;
          f2 = tt;
          count = max(f2, count);
        }
      }
    }
    count = max(f2, count);
    long long total = n - (2 * count);
    long long cool = min(ans, total);
    long long cool2 = min(cool, warna);
    if (cool2 < 0) {
      cool2 = 0;
    }
    cout << cool2 << "\n";
  }
  return 0;
}
