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>
template <class T>
T read(T &a) {
  a = 0;
  char x = getchar();
  bool f = 0;
  for (; x < '0' || x > '9'; x = getchar()) f |= x == '-';
  for (; x >= '0' && x <= '9'; x = getchar()) a = (a << 3) + (a << 1) + x - '0';
  if (f) a = -a;
  return a;
}
using namespace std;
const int N = 2e5 + 5;
char s[N];
int cnt[50];
int main() {
  int T;
  read(T);
  while (T--) {
    scanf("%s", s);
    int n = strlen(s);
    int ans = 0x3f3f3f3f;
    for (int i = 0; i <= 9; ++i) {
      cnt[i] = 0;
    }
    for (int i = 0; i < n; ++i) {
      cnt[s[i] - '0']++;
    }
    for (int i = 0; i <= 9; ++i) {
      ans = min(ans, n - cnt[i]);
    }
    for (int i = 0; i <= 9; ++i) {
      for (int j = 0; j <= 9; ++j) {
        if (i == j) continue;
        bool pf = 0;
        int ct = 0;
        for (int k = 0; k < n; ++k) {
          int num = s[k] - '0';
          if (num == i) {
            if (!pf) {
              pf = 1;
              ct++;
            }
          }
          if (num == j) {
            if (pf) {
              pf = 0;
              ct++;
            }
          }
        }
        ans = min(ans, n - (ct - pf));
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
