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 dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long expo(long long n, long long m, long long p) {
  long long r = 1;
  n = n % p;
  while (m > 0) {
    if (m % 2) r = (r * n) % p;
    n = (n * n) % p;
    m = m / 2;
  }
  return r % p;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long ans = (1LL << 61), n = s.size();
    for (long long i = 0; i <= 9; i++) {
      for (long long j = 0; j <= 9; j++) {
        long long c = 0, f = 0;
        for (long long k = 0; k < n; k++) {
          if (f % 2 == 0) {
            if (s[k] - '0' == i) f++, c++;
          } else {
            if (s[k] - '0' == j) f++, c++;
          }
        }
        if (c % 2 && i != j) c--;
        ans = min(ans, n - c);
      }
    }
    cout << ans << '\n';
  }
}
