human
Bob has a permutation of integers from 1 to n. Denote this permutation as p. The i-th element of p will be denoted as pi. For all pairs of distinct integers i, j between 1 and n, he wrote the number ai, j = min(pi, pj). He writes ai, i = 0 for all integer i from 1 to n.

Bob gave you all the values of ai, j that he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given.

Input

The first line of the input will contain a single integer n (2 ≤ n ≤ 50).

The next n lines will contain the values of ai, j. The j-th number on the i-th line will represent ai, j. The i-th number on the i-th line will be 0. It's guaranteed that ai, j = aj, i and there is at least one solution consistent with the information given.

Output

Print n space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print any of them.

Examples

Input

2
0 1
1 0

Output

2 1

Input

5
0 2 2 1 2
2 0 4 1 3
2 4 0 1 3
1 1 1 0 1
2 3 3 1 0

Output

2 5 4 1 3

Note

In the first case, the answer can be {1, 2} or {2, 1}.

In the second case, another possible answer is {2, 4, 5, 1, 3}.
Using python3 can you solve the prior task?

Give me a Python program to print n space separated integers, which represents a permutation that could have generated these values?
assistant
#Here is Python program to print n space separated integers, which represents a permutation that could have generated these values.

#!/usr/bin/env python3

try:
    while True:
        n = int(input())
        a = [list(map(int, input().split())) for i in range(n)]
        used = [False] * n
        result = [0] * n

        def is_good(i):
            if used[i]:
                return False
            prev = None
            for j, x in enumerate(a[i]):
                if not used[j] and i != j:
                    if prev is None:
                        prev = x
                    elif prev != x:
                        return False
            return True

        for step in range(n):
            first_good = next(filter(is_good, range(n)))
            used[first_good] = True
            for j, x in enumerate(a[first_good]):
                if not used[j]:
                    result[first_good] = x
                    break
            else:
                result[first_good] = n

        print(' '.join(map(str, result)))

except EOFError:
    pass