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;
double EPS = 1e-9;
long long INF = 1000000000000000005ll;
double PI = acos(-1);
inline string IntToString(long long int a) {
  char x[100];
  sprintf(x, "%lld", a);
  string s = x;
  return s;
}
inline long long int StringToInt(string a) {
  char x[100];
  long long int res;
  strcpy(x, a.c_str());
  sscanf(x, "%lld", &res);
  return res;
}
inline string uppercase(string s) {
  int n = (int)s.size();
  for (long long int i = 0; i < n; i++)
    if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}
inline string lowercase(string s) {
  int n = (int)s.size();
  for (long long int i = 0; i < n; i++)
    if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  long long int as = abs(a.first - a.second);
  long long int binary_search = abs(b.first - b.second);
  return (as > binary_search);
}
long long int dx[] = {-1, 0, 1, 0};
long long int dy[] = {0, 1, 0, -1};
vector<long long int> v[100005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int n = s.length();
    long long int a[10] = {0};
    for (long long int i = 0; i < n; i++) {
      a[s[i] - '0']++;
    }
    long long int ma = INT_MIN;
    for (long long int i = 0; i < 10; i++) {
      ma = max(ma, a[i]);
    }
    for (long long int i = 0; i < 10; i++) {
      for (long long int j = 0; j < 10; j++) {
        if (j != i) {
          long long int cnt = 0;
          bool x = 0;
          for (long long int k = 0; k < n; k++) {
            if (!x) {
              if (s[k] - '0' == i) {
                x = 1;
                cnt++;
              }
            } else {
              if (s[k] - '0' == j) {
                cnt++;
                x = 0;
              }
            }
          }
          if (cnt % 2 == 0)
            ma = max(ma, cnt);
          else
            ma = max(ma, cnt - 1);
        }
      }
    }
    cout << n - ma << '\n';
  }
  return 0;
}
