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 int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 50;
int n, cnt[N], dp[N][2];
char s[N];
void work() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = '0'; i <= '9'; ++i) cnt[i] = 0;
  for (int i = 1; i <= n; ++i) ++cnt[s[i]];
  int ans = 0;
  for (int i = '0'; i <= '9'; ++i) {
    int ret = cnt[i];
    ans = max(ans, ret);
  }
  for (int i = '0'; i <= '9'; ++i)
    for (int j = '0'; j <= '9'; ++j)
      if (i != j) {
        dp[0][0] = -INF;
        dp[0][1] = -INF;
        for (int k = 1; k <= n; ++k) {
          dp[k][0] = dp[k - 1][0];
          dp[k][1] = dp[k - 1][1];
          if (s[k] == i) dp[k][0] = max({1, dp[k][0], dp[k - 1][1] + 1});
          if (s[k] == j) dp[k][1] = max(dp[k][1], dp[k - 1][0] + 1);
        }
        ans = max(ans, dp[n][1]);
      }
  printf("%d\n", n - ans);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) work();
}
