E. Dasha and Puzzle（dfs+思维脑洞）

Dasha decided to have a rest after solving the problem. She had been ready to start her favourite activity — origami, but remembered the puzzle that she could not solve.

The tree is a non-oriented connected graph without cycles. In particular, there always are n - 1 edges in a tree with n vertices.

The puzzle is to position the vertices at the points of the Cartesian plane with integral coordinates, so that the segments between the vertices connected by edges are parallel to the coordinate axes. Also, the intersection of segments is allowed only at their ends. Distinct vertices should be placed at different points.

Help Dasha to find any suitable way to position the tree vertices on the plane.

It is guaranteed that if it is possible to position the tree vertices on the plane without violating the condition which is given above, then you can do it by using points with integral coordinates which don't exceed 10¹⁸ in absolute value.

Input

The first line contains single integer n (1 ≤ n ≤ 30) — the number of vertices in the tree.

Each of next n - 1 lines contains two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) that mean that the i-th edge of the tree connects vertices u_i and v_i.

It is guaranteed that the described graph is a tree.

Output

If the puzzle doesn't have a solution then in the only line print "NO".

Otherwise, the first line should contain "YES". The next n lines should contain the pair of integers x_i, y_i (|x_i|, |y_i| ≤ 10¹⁸) — the coordinates of the point which corresponds to the i-th vertex of the tree.

If there are several solutions, print any of them.

Examples

Input

7
1 2
1 3
2 4
2 5
3 6
3 7

Output

YES
0 0
1 0
0 1
2 0
1 -1
-1 1
0 2

Input

6
1 2
2 3
2 4
2 5
2 6

Output

NO

Input

4
1 2
2 3
3 4

Output

YES
3 3
4 3
5 3
6 3

Note

In the first sample one of the possible positions of tree is:

题意： 就是给你N-1条边，让你在这个完全的坐标轴上，看看是否形成一个树，每一个点都必须在不同的位置，这些边只能在结尾的位置相交。这些边必须和坐标走平行。

思路：我们现在思考一下，这些边什么时候不符合情况，就是有有一个顶点，与他相连的节点数大于4，这个时候是肯定不可以的，是不是剩下的情况就是可以的啊，答案是肯定的，你们你们不懂的可以在纸上画一画，我们只要先选择一个1为根的话，然后dfs下去，我们每次选一个顶点，然后每次选的距离都是2^n这样的话我们肯定是不会有相交的边的的情况的了。所以我们 现在要做的就是判断一下不符合情况的，然后再计算就可以了，其实就是一个dfs，我们只要选择第一个点为起点就可以了，时间复杂度，好像就是O(n);  我们每次都让距离减半这样子的话就没有相交的点了，感觉不是那么的难这种问题以后在纸上画画就行了。