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 long double PI = atan(1.0) * 4;
const int64_t INF = 3e18;
const int32_t LOG = 21;
const int32_t MOD = 1e9 + 7;
void files_init() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int tests = 1;
  cin >> tests;
  for (int ii = 0; ii < tests; ii++) {
    string second;
    cin >> second;
    vector<int> v;
    vector<int> cnt(10, 0);
    for (auto it : second) v.push_back(it - '0');
    int n = v.size();
    for (auto it : v) cnt[it]++;
    int max1 = 0;
    for (int i = 0; i < 10; i++) max1 = max(max1, cnt[i]);
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        if (i == j) continue;
        int len = 0;
        int check = i;
        for (auto& it : v) {
          if (check == it) {
            len++;
            check = i + j - check;
          }
        }
        if (len & 1) len--;
        max1 = max(max1, len);
      }
    }
    cout << n - max1 << '\n';
  }
  return 0;
}
