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;
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
const int MOD = 998244353, N = 2e5 + 5;
int a[11], dp[11][11];
int main() {
  cin.sync_with_stdio(false), cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    fill(a, a + 10, 0);
    string s;
    cin >> s;
    n = s.size();
    for (int c = 0; c < 10; c++) {
      for (int j = c + 1; j < 10; j++) {
        dp[c][j] = 0;
      }
    }
    for (int i = 0; i < s.size(); i++) a[s[i] - 48]++;
    int mx = *max_element(a, a + 10);
    int ls = -1;
    int tp = 0;
    for (int c = 0; c < 10; c++) {
      for (int j = c + 1; j < 10; j++) {
        tp = 0;
        ls = -1;
        for (int i = 0; i < n; i++) {
          if (ls == -1 && s[i] == c + 48) {
            ls = 2;
            tp++;
          } else if (ls == -1 && s[i] == j + 48) {
            ls = 1;
            tp++;
          } else if (ls == 1 && s[i] == c + 48) {
            ls = 2;
            tp++;
          } else if (ls == 2 && s[i] == j + 48) {
            ls = 1;
            tp++;
          }
        }
        dp[c][j] = tp;
      }
    }
    int ans, mxd = 0;
    for (int c = 0; c < 10; c++) {
      for (int j = c + 1; j < 10; j++) {
        mxd = max(mxd, dp[c][j]);
      }
    }
    for (int i = s.size(); i >= 2; i--) {
      ans = i;
      if (mx >= i) {
        break;
      } else if (i % 2 == 0 && mxd >= i)
        break;
    }
    cout << s.size() - ans << "\n";
  }
  return 0;
}
