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 long long INF = 1e18 * 9;
const int N = 2e5 + 23;
const int M = 3003;
const int mod1 = 1e9 + 7;
const int mod = 1e9 + 7;
const int base = 131;
int n, m, k, ans, cnt, x, y, mid, sx = -1, sy = -1, mx, mi = INF;
int a[N], b[N];
bool vis[N];
string s, ss;
map<int, int> mp, mp1;
vector<int> ve, ve1;
bool cmp(int a, int b) { return a > b; }
int cheak(int sx, int sy) {
  ve.clear();
  int s1 = 0;
  for (int i = 0; i < n; i++) {
    if (!(a[i] == sx || a[i] == sy)) continue;
    if (ve.size() < 2) {
      if (!ve.size() && a[i] == sx) ve.push_back(a[i]);
      if (ve.size() == 1 && a[i] == sy) ve.push_back(a[i]);
    } else {
      if (a[i] == ve[ve.size() - 2]) ve.push_back(a[i]);
    }
  }
  s1 = ve.size();
  return (s1 & 1 ? s1 - 1 : s1);
}
void solve() {
  int pos = 0, s1 = 0, s2 = 0, ans = mod, z, f = 0;
  mx = 0;
  cin >> s;
  n = s.size();
  mp.clear();
  for (int i = 0; i < s.size(); i++) {
    a[i] = (s[i] - '0');
    mp[a[i]]++;
    ans = min(ans, n - mp[a[i]]);
  }
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      if (i == j) continue;
      ans = min(ans, n - cheak(i, j));
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int T;
  for (cin >> T; T--;) solve();
}
