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;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int test;
  cin >> test;
  long long int get, ans, x, y;
  while (test--) {
    string str;
    cin >> str;
    long long int n = str.size();
    vector<long long int> v;
    x = 0;
    y = 0;
    long long int i, j, k;
    for (i = 0; i < 10; i++) {
      for (j = i + 1; j < 10; j++) {
        x = 0;
        get = 0;
        y = 0;
        ans = 0;
        for (k = 0; k < n; k++) {
          if (str[k] - '0' == j) y = 1;
          if (str[k] - '0' == i and y == 1) {
            ans++;
            y = 0;
          }
          if (str[k] - '0' == i) x = 1;
          if (str[k] - '0' == j and x == 1) {
            get++;
            x = 0;
          }
        }
        ans = (2 * ans);
        get = (2 * get);
        v.push_back(get);
        v.push_back(ans);
      }
    }
    x = 0;
    y = 0;
    long long int akhilesh[10] = {};
    for (i = 0; i < n; i++) {
      x = str[i] - '0';
      akhilesh[x]++;
    }
    x = 0;
    for (i = 0; i < 10; i++) x = max(x, akhilesh[i]);
    get = 2;
    x = max(get, x);
    for (auto p : v) y = max(y, p);
    get = max(x, y);
    cout << n - get << "\n";
  }
  return 0;
}
