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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
inline void write(int x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
inline void writeputs(int x) { write(x), putchar('\n'); }
inline void writeputchar(int x) { write(x), putchar(' '); }
int t, n, num[15], dp[10][10], ANS;
char s[200005];
int main() {
  t = read();
  while (t--) {
    memset(num, 0, sizeof num);
    memset(dp, 0, sizeof dp);
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    ANS = 0;
    for (int i = 1; i <= len; i++) num[s[i] - '0']++;
    for (int i = 1; i <= len; i++) {
      int ch = s[i] - '0';
      for (int j = 0; j <= 9; j++) {
        if (j == ch) continue;
        if (dp[j][ch] % 2 != 0) dp[j][ch]++;
        if (dp[ch][j] % 2 == 0) dp[ch][j]++;
      }
    }
    for (int i = 0; i <= 9; i++) ANS = max(ANS, num[i]);
    for (int i = 0; i <= 9; i++)
      for (int j = 0; j <= 9; j++)
        if (i != j) ANS = max(ANS, dp[i][j] / 2 * 2);
    writeputs(len - ANS);
  }
  return 0;
}
