#include <bits/stdc++.h>
using namespace std;
int gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long i, j, k, n, m, t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    n = s.size();
    map<char, long long> make_pair;
    long long mx = INT_MIN;
    for (i = 0; i < n; i++) {
      make_pair[s[i]]++;
    }
    for (char c = '0'; c <= '9'; c++) {
      if (make_pair[c] > mx) {
        mx = make_pair[c];
      }
    }
    for (char c1 = '0'; c1 < '9'; c1++) {
      for (char c2 = c1 + 1; c2 <= '9'; c2++) {
        k = 0;
        char prev = '$';
        for (i = 0; i < n; i++) {
          if (s[i] == c1) {
            if (prev == c2) {
              prev = c1;
              k++;
            } else if (prev == '$') {
              prev = c1;
              k++;
            }
          } else if (s[i] == c2) {
            if (prev == c1) {
              prev = c2;
              k++;
            } else if (prev == '$') {
              prev = c2;
              k++;
            }
          }
        }
        if (k > mx && k % 2 == 0) {
          mx = k;
        } else if ((k - 1) > mx && (k - 1) % 2 == 0) {
          mx = k - 1;
        }
      }
    }
    cout << s.size() - mx << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long i, j, k, n, m, t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    n = s.size();
    map<char, long long> make_pair;
    long long mx = INT_MIN;
    for (i = 0; i < n; i++) {
      make_pair[s[i]]++;
    }
    for (char c = '0'; c <= '9'; c++) {
      if (make_pair[c] > mx) {
        mx = make_pair[c];
      }
    }
    for (char c1 = '0'; c1 < '9'; c1++) {
      for (char c2 = c1 + 1; c2 <= '9'; c2++) {
        k = 0;
        char prev = '$';
        for (i = 0; i < n; i++) {
          if (s[i] == c1) {
            if (prev == c2) {
              prev = c1;
              k++;
            } else if (prev == '$') {
              prev = c1;
              k++;
            }
          } else if (s[i] == c2) {
            if (prev == c1) {
              prev = c2;
              k++;
            } else if (prev == '$') {
              prev = c2;
              k++;
            }
          }
        }
        if (k > mx && k % 2 == 0) {
          mx = k;
        } else if ((k - 1) > mx && (k - 1) % 2 == 0) {
          mx = k - 1;
        }
      }
    }
    cout << s.size() - mx << endl;
  }
  return 0;
}

```