ADACYCLE - Ada and Cycle

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Ada the Ladybug is on a trip in Bugindia. There are many cities and some uni-directional roads connecting them. Ada is wondering about the shortest path, which begins in a city and ends in the same city. Since Ada likes short trips, she asked you to find the length of such path for each city in Bugindia.

Input

The first line will contain 0 < N ≤ 2000, the number of cities.

Then N lines follow, each containing N integers 0 ≤ Hij ≤ 1. One means, that there is a road between i and j (zero means there isn't a road).

Output

Print N lines, the length of shortest path which begins in city i and ends in city i. If the path doesn't exist, print "NO WAY" instead.

Example Input

5
0 1 1 1 1
1 0 0 0 1
0 0 1 1 0
0 0 1 0 0
0 0 0 1 0

Example Output

2
2
1
2
NO WAY

Example Input

5
0 1 0 0 1
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1 0 0 0 0

Example Output

2
5
5
5
2

hide comments
morass: 2016-09-15 11:54:25

@Harsh Rumalwala: Good day to you! Well In my opinion, your algorithm is O(N*E+N^2) [E is number of edges], where E might be N^2 in worst-case ... so it is O(N^3) in worst case. Isn't that so? :O

Harsh Rumalwala: 2016-09-15 07:19:28

Why O(n^2) solution is resulting in TLE while it should easily pass under the given time constraints?

morass: 2016-09-13 16:05:56

@[Rampage] Blue.Mary: Good day to you man! Nope, it is expected worst complexity "with some improvisation" if you understand me :)

[Rampage] Blue.Mary: 2016-09-13 16:00:29

What's the expected complexity? Is a (strictly) O(n^2) algorithm required?

[Edited After AC] It only requires some observation.

Last edit: 2016-09-18 13:04:31

Added by:Morass
Date:2016-09-13
Time limit:3s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64 GOSU