The following languages are only available languages for the problems from the contest

Surprise Language Round #8:

- Kotlin 1.0.5-2

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

I. Loader

time limit per test

4 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputA loader works in a warehouse, which is a rectangular field with size *n* × *m*. Some cells of this field are free, others are occupied by pillars on which the roof of the warehouse rests.

There is a load in one of the free cells, and the loader in another. At any moment, the loader and the load can not be in the cells with columns, outside the warehouse or in the same cell.

The loader can move to the adjacent cell (two cells are considered adjacent if they have a common side), or move the load. To move the load, the loader should reach the cell adjacent to the load and push the load. In this case the load advances to the next cell in the direction in which the loader pushes it and the loader ends up in the cell in which the load was.

Your task is to determine a sequence of pushes and loader's movements after which the load will reach the given cell (it is guaranteed that this cell is free). The load is rather heavy, so you need to minimize first the number of pushes and second the number of loader's movements.

Input

The first line contains two positive integers *n* and *m* (1 ≤ *n*, *m* ≤ 40, *n*·*m* ≥ 3) — the number of rows and columns in the rectangular field.

Each of the next *n* lines contains *m* characters — the description of the warehouse. If there is a character in the next cell of the warehouse:

- "X", it means, that the current cell contains the column;
- ".", it means, that the current cell is free;
- "Y", it means, that the loader is in the current cell;
- "B", it means, that the load is in the current cell;
- "T", it means, that the load should be moved to this cell.

It is guaranteed that there is exactly one load, one loader and one cell to which the load should be moved.

Output

If the loader is not able to move the load to the given cell, print "NO" (without the quotes) in the first line of the output.

Otherwise, print "YES" (without the quotes) in the first line of the output, and in the second line — the sequence of characters that determines movements and pushes of the loader. Characters w, e, n, s shall denote loader's moves to the west, east, north and south, respectively. Characters W, E, N, S must denote loader's pushes in the corresponding directions. First of all you need to minimize the number of pushes of the load and second, the number of movements of the loader. If there are several answers, you are allowed to print any of them.

Examples

Input

3 3

..Y

.BX

..T

Output

YES

wSwsE

Input

3 3

.BY

...

TXX

Output

NO

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Oct/17/2017 09:09:23 (c4).

Desktop version, switch to mobile version.

User lists

Name |
---|