Guess the Permutation

Description

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.

Sample Input

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

Sample Output

Hint

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}.

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;

int main()
{
int n, i, j, m;
int a[50][50], b[50];
scanf("%d", &n);
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
scanf("%d", &a[i][j]);
}
getchar();
}
for (i = 0; i < n; i++)
{
m = 0;
for (j = 1;j < n; j++)
{
if (a[i][m] < a[i][j])
m = j;
}
b[i] = a[i][m];
if (b[i] > n || b[i] == n - 1)
b[i] = n;
}
printf("%d", b[0]);
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
if (b[i] == b[j])
b[j] -= 1;
}
if (i != 0)
{
printf(" %d", b[i]);
}

}
return 0;
}