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("Ofast")
using namespace std;
long long binpow(long long base, long long exp, long long mod) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % mod;
    exp = exp >> 1;
    base = (base * base) % mod;
  }
  return res;
}
long long mod(long long x) {
  return ((x % 1000000007LL + 1000000007LL) % 1000000007LL);
}
long long add(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mul(long long a, long long b) { return mod(mod(a) * mod(b)); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long LongestEvenSequence = 0;
    long long MostFrequentCount = 0;
    unordered_map<long long, long long> frq;
    vector<vector<long long> > dp;
    dp.resize(10, vector<long long>(10, 0));
    for (long long i = 0; i < s.length(); i++) {
      long long val = s[i] - '0';
      frq[val]++;
      for (long long j = 0; j <= 9; j++) {
        dp[val][j] = dp[j][val] + 1;
        if (dp[val][j] % 2 == 0) {
          LongestEvenSequence = max(LongestEvenSequence, dp[val][j]);
        }
      }
    }
    for (auto x : frq) {
      MostFrequentCount = max(x.second, MostFrequentCount);
    }
    cout << min(s.length() - MostFrequentCount,
                s.length() - LongestEvenSequence)
         << "\n";
    ;
  }
  return 0;
}
