21641_BoredS

One day, S was so bored that he did something boring like this.

He drew N points on the paper, and added a loop, that is, an edge connecting a point and itself, on each point. Then, he added several edges between different points, randomly. It's guaranteed that there was no more than one edge between a pair of points. So, there he gained an undirected graph G without any parallel edge. Finally, he figured out the adjacent matrix A of the graph G, and calculated A².

Here you're given only the finally matrix A², S wants to know whether there is a way to travel all the edges on graph G once and exactly once, and end at the point where the journey starts. Could you answer him?

The first line is an integer T ( T ≤ 20 ), indicating the number of test cases. Each case is in the format like this:

First line is a positive integer N ( N ≤ 100 ), indicating the number of points of the graph. Then there are N lines following, with N space-separated integers on each line. The j-th integer on the i-th line is the (i, j) element of matrix A².

For each test case, print one line with the string "Yes" or "No" (quotes for clarity), according to your answer to the question above.

2
1
1
2
2 2
2 2

Yes
No

B = A²

if and only if

( A is an n*n matrix and the elements are 0-based )