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 int;
const ll mod = 1000000007;
const ll M = 1000000000000000000ll;
inline ll add(ll a, ll b) { return ((a % mod + b % mod) % mod); }
inline ll sub(ll a, ll b) { return ((a % mod - b % mod + mod) % mod); }
inline ll mult(ll a, ll b) { return (((a % mod) * (b % mod)) % mod); }
inline ll power(ll n, ll x) {
  ll p = 1;
  while (x != 0) {
    if (x % 2 == 0) {
      n = mult(n, n);
      x = x / 2;
    }
    p = mult(p, n);
    x--;
  }
  return p;
}
inline ll recurmult(ll a, ll b) {
  ll r = 0;
  while (b) {
    if (b & 1) r = add(r, a);
    a = mult(2, a);
    b >>= 1;
  }
  return r % mod;
}
ll findans(set<ll>& s1, set<ll>& s2) {
  ll ans = 1;
  if (s1.size() == 0 || s2.size() == 0) return ans;
  ll x1 = *s1.begin();
  ll x2 = *s2.begin();
  if (x1 < x2) {
    while (1) {
      auto it1 = s2.upper_bound(x1);
      if (it1 == s2.end()) break;
      ans++;
      auto it2 = s1.upper_bound(*it1);
      if (it2 == s1.end()) break;
      x1 = *it2;
      ans++;
    }
  } else {
    while (1) {
      auto it1 = s1.upper_bound(x1);
      if (it1 == s1.end()) break;
      ans++;
      auto it2 = s2.upper_bound(*it1);
      if (it2 == s2.end()) break;
      x1 = *it2;
      ans++;
    }
  }
  return (ans / 2) * 2;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    set<ll> s1[10];
    for (ll i = 0; i < s.length(); i++) {
      s1[s[i] - '0'].insert(i);
    }
    ll ans = 0;
    for (ll i = 0; i < 10; i++) {
      for (ll j = 0; j < 10; j++) {
        if (i != j)
          ans = max(ans, findans(s1[i], s1[j]));
        else
          ans = max(ans, (ll)s1[i].size());
      }
    }
    cout << s.length() - ans << "\n";
  }
}
