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 int N = 2E5 + 10;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> s;
    int n = s.length();
    int cnt[10] = {0};
    for (int i = 0; i <= n - 1; i++) cnt[s[i] - '0']++;
    sort(cnt, cnt + 10);
    if (cnt[9] == n) {
      cout << 0 << endl;
      continue;
    }
    int ans = n - cnt[9];
    int ct[10][10] = {0}, add[10][10] = {0};
    for (int i = 0; i <= n - 1; i++) {
      int x = s[i] - '0';
      for (int k = 0; k <= 9; k++)
        if (k != x) {
          if (ct[x][k] == 0) ct[x][k]++;
          if (ct[k][x] == 1) {
            ct[k][x]--;
            add[k][x]++;
          }
        }
    }
    for (int i = 0; i <= 9; i++)
      for (int j = 0; j <= 9; j++) ans = min(ans, n - add[i][j] * 2);
    cout << ans << endl;
  }
}
