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.
for t in range(int(input())):
    s=input()
    n=len(s)
    index=[[] for i in range(10)]
    for i in range(len(s)):
        index[int(s[i])].append(i)
    ans=float('inf')
    #print(index)
    for i in range(10):
        for j in range(10):
            if i==j:
                ans=min(ans,n-len(index[i]))
            else:
                i1=j1=0
                cnt=0
                last=-1
                #arr=[]
                while True:
                    if cnt%2==0:
                        if i1==len(index[i]):break
                        if index[i][i1]>last:
                            last=index[i][i1]
                            #arr.append(index[i][i1])
                            cnt+=1
                        i1+=1
                    else:
                        if j1==len(index[j]):break
                        if index[j][j1]>last:
                            last=index[j][j1]
                            #arr.append(index[j][j1])
                            cnt+=1
                        j1+=1
                if cnt%2:cnt-=1
                ans=min(ans,n-cnt)
                #if cnt>0:print(i,j,cnt,arr,ans)
    print(ans)