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.
t = int(input())

for _ in range(t):
    #n = int(input())
    s = input()
    l = [0]*10
    for i in range(10):
        l[i] = s.count(str(i))
    
    #print(l)

    minans = len(s) - max(l)

    ans = -1
##    for i in range(10):
##        v = i
##        last = s.find(str(v))
##        ll = last
##        if last!=-1:
##            for j in range(10):
##                last = s.find(str(j),ll)
##                if last!= -1:
##                    v = j
##                #print(i, j)
##                if i == j:
##                    continue
##                kol = 1
##                while last!=-1:
##                    kol += 1
##                    if v == i:
##                        v = j
##                    else:
##                        v = i
##                    last = s.find(str(v), last)
##                print(i, j, kol)
##                ans = max(ans, kol)



    for i in range(10):
        if s.find(str(i)) != -1:
            last = s.find(str(i))
            ll =last
            for j in range(10):
                last = ll
                #print(i, "at", last)
                kol = 0
                if s.find(str(j), last+1) != -1:
                    kol = 2
                    last = s.find(str(j), last+1)
                    #print(j, "at", last)
                    ok = True
                    while ok:
                        if s.find(str(i), last+1) != -1:
                            last = s.find(str(i), last+1)
                            #print(i, "at", last)
                            if s.find(str(j), last+1) != -1:
                                last = s.find(str(j), last+1)
                                #print(j, "at", last)
                                kol += 2
                            else: ok = False
                        else: ok = False
                #print(kol)
                ans = max(ans, kol)
                #print(ans)
        
    print(min(minans, len(s)-ans))


















    
