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;
const long long int llINF = 0x3f3f3f3f3f3f3f;
int main() {
  int t;
  cin >> t;
  while (t--) {
    vector<int> pos[10];
    string s;
    cin >> s;
    int n = s.size();
    for (int i = 0; i < n; i++) {
      pos[s[i] - '0'].push_back(i);
    }
    int ans = -1;
    for (int i = 0; i < 10; i++) ans = max(ans, (int)pos[i].size());
    for (int i = 0; i < 10; i++) {
      if (pos[i].size() == 0) continue;
      for (int j = 0; j < 10; j++) {
        if (i == j) continue;
        if (pos[j].size() == 0) continue;
        int ni = pos[i].size(), nj = pos[j].size();
        int md = 0, curi = 0, curj = 0, last = -1;
        int plc = 0;
        while (curi < ni && curj < nj) {
          while (!md && curi < ni && pos[i][curi] < last) curi++;
          if (!md && curi < ni) {
            last = pos[i][curi];
            plc++;
            md ^= 1;
            continue;
          }
          while (md && curj < nj && pos[j][curj] < last) curj++;
          if (md && curj < nj) {
            last = pos[j][curj];
            md ^= 1;
            plc++;
            continue;
          }
        }
        ans = max(ans, (plc / 2) * 2);
      }
    }
    printf("%d\n", n - ans);
  }
}
