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;
namespace Qza {
char s[200005];
int buc[256];
int max(int a, int b) { return a < b ? b : a; }
void main() {
  int t, n, m, p, ans;
  bool flag;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s);
    n = strlen(s);
    ans = 0;
    for (int i = '0'; i <= '9'; ++i) buc[i] = 0;
    for (int i = 0; i < n; ++i) ++buc[s[i]];
    for (int i = '0'; i <= '9'; ++i) ans = max(ans, buc[i]);
    for (int i = '1'; i <= '9'; ++i) {
      for (int j = '0'; j < i; ++j) {
        if (i == j) continue;
        p = 0;
        m = 0;
        flag = true;
        while (p < n) {
          if (flag) {
            while (p < n && s[p] != i) ++p;
            if (p < n) ++m, flag = false;
          } else {
            while (p < n && s[p] != j) ++p;
            if (p < n) ++m, flag = true;
          }
        }
        m -= (m & 1);
        ans = max(ans, m);
        p = 0;
        m = 0;
        flag = false;
        while (p < n) {
          if (flag) {
            while (p < n && s[p] != i) ++p;
            if (p < n) ++m, flag = false;
          } else {
            while (p < n && s[p] != j) ++p;
            if (p < n) ++m, flag = true;
          }
        }
        m -= (m & 1);
        ans = max(ans, m);
      }
    }
    printf("%d\n", n - ans);
  }
}
}  // namespace Qza
int main() {
  Qza::main();
  return 0;
}
