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;
using ll = long long;
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
int power(int a, int b) {
  if (!b) return 1;
  int c = power(a, b / 2);
  c = (1LL * c * c) % mod;
  if (b % 2) c = (1LL * c * a) % mod;
  return c;
}
void done() {}
int dp[10][10];
void solve() {
  for (int i = 0; i < 10; ++i) {
    for (int j = 0; j < 10; ++j) {
      dp[i][j] = 0;
    }
  }
  string s;
  cin >> s;
  int n = s.size();
  for (char c : s) {
    for (int i = 0; i < 10; ++i) {
      dp[c - '0'][i] = dp[i][c - '0'] + 1;
    }
  }
  int res = 0;
  for (int i = 0; i < 10; ++i) {
    for (int j = 0; j < 10; ++j) {
      if (i == j) {
        res = max(res, dp[i][j]);
      } else {
        if (dp[i][j] & 1) {
          res = max(res, dp[i][j] - 1);
        } else {
          res = max(res, dp[i][j]);
        }
      }
    }
  }
  cout << n - res << '\n';
}
void another() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<int> r(n, 1), l(n, 1);
  int ans = 0;
  for (int i = n - 2; i >= 0; --i) {
    if (a[i + 1] > a[i]) r[i] = r[i + 1] + 1;
    ans = max(ans, r[i]);
  }
  for (int i = 1; i < n; ++i) {
    if (a[i - 1] < a[i + 1]) {
      l[i] = l[i - 1] + 1;
    }
  }
  for (int i = 0; i < n - 2; ++i) {
  }
  cout << ans << '\n';
}
void test_case() {
  int t;
  cin >> t;
  while (t--) solve();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  test_case();
}
