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;
using LL = long long;
int T;
int N, K, L, R;
int P[1000100], M[1000100];
int D[1000100];
int A[1000100], B[1000100];
char S[1000100];
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s", S);
    int ans = 0;
    int len = strlen(S);
    map<char, int> cnt;
    for (int i = 0; i < len; ++i) cnt[S[i]]++;
    for (auto& p : cnt) ans = max(ans, p.second);
    for (char a = '0'; a <= '9'; ++a) {
      for (char b = '0'; b <= '9'; ++b) {
        if (a == b || cnt.count(a) == 0 || cnt.count(b) == 0) continue;
        int idx = 0;
        int this_cnt = 0;
        for (int i = 0; i < len; ++i) {
          if (idx == 0 && S[i] != a || idx == 1 && S[i] != b) continue;
          idx = 1 - idx;
          this_cnt++;
        }
        ans = max(ans, this_cnt / 2 * 2);
      }
    }
    printf("%d\n", len - ans);
  }
  return 0;
}
