codeforces618B

Guess the Permutation

CodeForces - 618B

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 p_i. For all pairs of distinct integers i, jbetween 1 and n, he wrote the number a_i, j = min(p_i, p_j). He writes a_i, i = 0 for all integer i from 1 to n.

Bob gave you all the values of a_i, 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 a_i, j. The j-th number on the i-th line will represent a_i, j. The i-th number on the i-th line will be 0. It's guaranteed that a_i, j = a_j, 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}.

sol：容易发现对于有且仅有一个序列只有1，有且仅有2个序列只有1,2······

这样就可以对每行维护一个前缀和表示该行数字1~i的数字出现的次数和，然后从1到n一个个数字填过去即可

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+'');    return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n;
int a[N][N],Ges[N][N];
int Ans[N];
bool Arr[N];
int main()
{
    int i,j;
    R(n);
    for(i=;i<=n;i++)
    {
        for(j=;j<=n;j++)
        {
            R(a[i][j]);
            Ges[i][a[i][j]]++;
        }
        for(j=;j<=n;j++) Ges[i][j]=Ges[i][j-]+Ges[i][j];
    }
    for(i=;i<=n;i++)
    {
        for(j=;j<=n;j++) if((Ges[j][i]==n-)&&(!Arr[j]))
        {
            Ans[j]=i;
            Arr[j]=;
            break;
        }
    }
    for(i=;i<=n;i++) W(Ans[i]);
    return ;
}
/*
input
2
0 1
1 0
output
2 1 或 1 2 

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 或 2 4 5 1 3
*/