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);
  long long i, j, k, a, b, c, t, x, y, n, m, z, flag = 0, ans = 0, sum = 0;
  cin >> t;
  while (t--) {
    string s1, s2, st;
    cin >> st;
    n = st.length();
    vector<long long> v(10, 0);
    for (i = 0; i < n; i++) {
      v[st[i] - '0']++;
    }
    ans = LONG_LONG_MIN;
    for (i = 0; i < 10; i++) {
      for (j = i + 1; j < 10; j++) {
        if (v[i] == 0 || v[j] == 0) continue;
        a = 0;
        x = 0;
        for (k = 0; k < n; k++) {
          z = st[k] - '0';
          if (z == i)
            a++;
          else if (z == j) {
            if (a != 0) {
              x++;
              a = 0;
            }
          }
        }
        ans = max(ans, x);
        a = 0;
        x = 0;
        for (k = 0; k < n; k++) {
          z = st[k] - '0';
          if (z == j)
            a++;
          else if (z == i) {
            if (a != 0) {
              x++;
              a = 0;
            }
          }
        }
        ans = max(ans, x);
      }
    }
    long long ans1 = LONG_LONG_MAX;
    if (ans != LONG_LONG_MIN) ans1 = n - (2 * ans);
    for (i = 0; i < 10; i++) {
      x = n - v[i];
      ans1 = min(ans1, x);
    }
    cout << ans1 << "\n";
  }
  return 0;
}
