【Wunder Fund Round 2016 (Div 1 + Div 2 combined)B】【暴力 贪心】Guess the Permutation 全排列a[i][j]=min(p[i],p

B. Guess the Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, j between 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 be0. 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

20 11 0

output

2 1

input

50 2 2 1 22 0 4 1 32 4 0 1 31 1 1 0 12 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}.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int n;int a[55][55];int val[55];int main(){while (~scanf("%d", &n)) {for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n;++j){scanf("%d", &a[i][j]);}}MS(val, 0);for (int v = 1; v <= n; ++v){for (int i = 1; i <= n; ++i)if(val[i]==0){bool flag = 1;for (int j = 1; j <= n; ++j){if (a[i][j] > v)flag = 0;}if (flag){val[i] = v;break;}}}for (int i = 1; i <= n; ++i)printf("%d ", val[i]);puts("");}return 0;}/*【题意】给你数字n（2<=n<=50），然后给你一个长度为n的全排列p[]。然后a[i][j]=min(p[i],p[j])，i!=j如果i==j，那么a[i][j]=0给出你a[][]，让你求p[]【类型】暴力【分析】我们可以依次从小到大枚举，依次找到1,2,3,...所有数在第几行。【时间复杂度&&优化】O（n^3）*/