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 inf = 0x3f3f3f3f;
long long mod = 998244353;
vector<long long> fact;
bool remender(long long a, long long b) { return a % b; }
long long add(long long x, long long y) { return (x + y) % mod; }
long long multi(long long x, long long y) { return (x * y) % mod; }
void fac(int n) {
  fact.push_back(1);
  for (long long i = 1; i <= n; i++) {
    fact.push_back(multi(i, fact.back()));
  }
}
long long power(long long x, long long y) {
  x %= mod;
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = multi(res, x);
    y = y >> 1;
    x = multi(x, x);
  }
  return res;
}
long long inverse(long long n) { return power(n, mod - 2); }
long long nCr(int n, int r) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (multi(fact[n], multi(inverse(fact[r]), inverse(fact[n - r])))) % mod;
}
void solve() {
  string s;
  cin >> s;
  int n = s.size();
  int ans = 0;
  int arr[10];
  memset(arr, 0, sizeof arr);
  for (int i = 0; i < n; i++) arr[s[i] - '0']++;
  for (int i = 0; i < 10; i++) {
    ans = max(ans, arr[i]);
  }
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      if (i == j) continue;
      int cnt = 0, l = i;
      for (int m = 0; m < n; m++) {
        if (s[m] - '0' == l) {
          cnt++;
          if (l == i)
            l = j;
          else
            l = i;
        }
      }
      if (cnt & 1) cnt--;
      ans = max(ans, cnt);
    }
  }
  cout << n - ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
