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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const int maxn = 50 * 50 + 10;
const int mod = (int)1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
int n;
int cnt[20];
char s[N];
void work() {
  cin >> s + 1;
  n = strlen(s + 1);
  memset(cnt, 0, sizeof(cnt));
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    s[i] -= '0';
    mx = max(mx, ++cnt[s[i]]);
  }
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      int cnt = 0;
      int x = i, y = j;
      for (int k = 1; k <= n; k++) {
        if (s[k] == x) {
          cnt++;
          swap(x, y);
        }
      }
      if (cnt & 1) cnt--;
      mx = max(mx, cnt);
    }
  }
  cout << n - mx << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) work();
  return 0;
}
