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.
# Contest No.: Edu92
# Problem No.: C
# Solver:      JEMINI
# Date:        20200729

import sys
import heapq

def main():
    t = int(input())
    for _ in range(t):
        strN = input()
        
        flag = False
        temp = strN[0]
        for i in strN:
            if i != temp:
                flag = True
        if not flag or len(strN) == 2:
            print(0)
            continue

        ans = len(strN)
        for i in ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]:
            for j in ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]:
                if j == i:
                    continue
                cnt = 0
                tempFlag = False
                for k in strN:
                    if k == i and not tempFlag:
                        tempFlag = True
                    elif k == j and tempFlag:
                        cnt += 2
                        tempFlag = False
                #print(i, j, cnt)
                ans = min(ans, len(strN) - cnt)
        
        for i in ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]:
            cnt = 0
            for k in strN:
                if k == i:
                    cnt += 1
            ans = min(ans, len(strN) - cnt)
        print(ans)

    return

if __name__ == "__main__":
    main()