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>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
char s[2000005];
int n, a[2000005], c[1221];
int calc(int u, int v) {
  int res = 0, flg = 0;
  for (register int i = (1); i <= (n); ++i) {
    if (!flg && a[i] == u) {
      res++;
      flg ^= 1;
    } else if (flg && a[i] == v) {
      res++;
      flg ^= 1;
    }
  }
  if (res & 1) res--;
  return res;
}
void work() {
  cin >> s + 1;
  n = strlen(s + 1);
  for (register int i = (0); i <= (9); ++i) c[i] = 0;
  for (register int i = (1); i <= (n); ++i) a[i] = s[i] - '0', c[a[i]]++;
  int res = 0;
  for (register int i = (0); i <= (9); ++i) res = max(res, c[i]);
  for (register int i = (0); i <= (9); ++i)
    for (register int j = (0); j <= (9); ++j)
      if (i != j) res = max(res, calc(i, j));
  cout << n - res << endl;
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}
