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 int64_t Nmax = 1e9 + 7;
const int64_t mod = 1000000007;
vector<int64_t> prime;
int64_t is_power_of_two(int64_t a) { return a && (!(a & (a - 1))); }
int64_t isperfectsquare(int64_t a) {
  double b = sqrt(a);
  return (b - ceil(b) == 0);
}
int64_t pow(int64_t a, int64_t b, int64_t m) {
  a = a % m;
  int64_t res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % m;
    }
    a = a * a % m;
    b = b / 2;
  }
  return res;
}
void seive(int64_t n) {
  bool isprime[n];
  memset(isprime, true, sizeof(isprime));
  prime.push_back(2);
  for (int64_t i = 3; i <= sqrt(n); i += 2) {
    if (isprime[i]) {
      for (int64_t j = i * i; j <= n; j += i) {
        isprime[j] = false;
      }
    }
  }
  for (int64_t i = 3; i <= n; i += 2) {
    if (isprime[i]) {
      prime.push_back(i);
    }
  }
}
void solve() {
  string s;
  cin >> s;
  int64_t f[10] = {0};
  int64_t mx = -1, mxx = -1;
  for (auto e : s) {
    f[e - '0']++;
  }
  for (auto e : f) {
    mx = max(mx, e);
  }
  mx = s.length() - mx;
  for (int64_t i = 0; i <= 9; i++) {
    for (int64_t j = 0; j <= 9; j++) {
      int64_t chg = 1;
      int64_t k = 0;
      int64_t rr = 0;
      if (i != j) {
        while (k < s.length()) {
          bool found = false;
          bool found1 = false;
          while (k < s.length() && (s[k] - '0') != i) {
            k++;
          }
          if (k < s.length() && (s[k] - '0') == i) {
            found = true;
          }
          while (k < s.length() && (s[k] - '0') != j) {
            k++;
          }
          if (k < s.length() && (s[k] - '0') == j) {
            found1 = true;
          }
          if (found1 && found) {
            rr++;
          }
        }
        mxx = max(mxx, rr);
      }
    }
  }
  cout << min(mx, (int64_t)s.length() - 2 * mxx) << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int64_t t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
