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;
void lc() { cout << endl; }
long long md(long long n) { return n % (long long)(1e9 + 7); }
string s;
vector<vector<long long> > dp;
long long calc(long long a, long long b, long long n) {
  long long r = 0;
  for (long long i = 0; i < n; i++) dp[i][0] = dp[i][1] = 0;
  for (long long i = 0; i < n; i++) {
    long long p = s[i] - '0';
    if (p == a) {
      if (i == 0)
        dp[i][0] = 1;
      else {
        dp[i][0] = dp[i - 1][1] + 1;
        dp[i][1] = dp[i - 1][1];
      }
    } else if (p == b) {
      if (i == 0)
        dp[i][1] = 1;
      else {
        dp[i][1] = dp[i - 1][0] + 1;
        dp[i][0] = dp[i - 1][0];
      }
    } else if (i > 0)
      dp[i][0] = dp[i - 1][0], dp[i][1] = dp[i - 1][1];
    r = max(r, max(dp[i][0], dp[i][1]));
  }
  if (r % 2) return n - r + 1;
  return n - r;
}
void solve() {
  cin >> s;
  long long n = s.size();
  long long cnt[10];
  memset(cnt, 0, sizeof(cnt));
  dp.resize(n);
  for (long long i = 0; i < n; i++) {
    dp[i].resize(2);
    cnt[s[i] - '0']++;
  }
  long long res = (long long)(1e18);
  for (long long i = 0; i < 10; i++) {
    for (long long j = i; j < 10; j++) {
      if (i == j)
        res = min(res, n - cnt[i]);
      else
        res = min(res, calc(i, j, n));
    }
  }
  cout << res;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(12) << fixed;
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
    lc();
  }
}
