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 nx = 2e5 + 5;
char mp[105][2], s[nx];
int t, n;
int cal(char a, char b) {
  int re = 0;
  if (a == b) {
    for (int i = 0; i < n; ++i) re += s[i] == a;
    return re;
  }
  bool f = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == a && !f) ++re, f = 1;
    if (s[i] == b && f) ++re, f = 0;
  }
  return re / 2 * 2;
}
int main() {
  int cnt = 0;
  for (int i = 0; i < 10; ++i)
    for (int j = 0; j < 10; ++j) mp[cnt][0] = '0' + i, mp[cnt++][1] = '0' + j;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s);
    n = strlen(s);
    int ans = 0;
    for (int i = 0; i < cnt; ++i) ans = max(ans, cal(mp[i][0], mp[i][1]));
    printf("%d\n", n - ans);
  }
}
