Problem: Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:

* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.

Constraints

* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.

Input

Input is given from Standard Input in the following format:


N M A B


Output

If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:


YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}


Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:

* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.

Examples

Input

3 4 4 2


Output

YES
<><>
^<>^
v<>v


Input

4 5 5 3


Output

YES
<>..^
^.<>v
v<>.^
<><>v


Input

7 9 20 20


Output

NO

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#define MN 1000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
char c[MN+5][MN+5];
int n,m,A,B;
int main()
{
	n=read();m=read();A=read();B=read();
	if(A*2+B*2>n*m) return 0*puts("NO");
	for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) c[i][j]='.';
	if(n&1) for(int j=2;j<=m&&A;j+=2) c[1][j-1]='<',c[1][j]='>',--A;
	if(m&1) for(int j=2;j<=n&&B;j+=2) c[j-1][m]='^',c[j][m]='v',--B;
	for(int j=m;j>1;--j) for(int i=1;i<=n;++i)
		if(c[i][j]=='.'&&c[i][j-1]=='.'&&A) --A,c[i][j]='>',c[i][j-1]='<';
	for(int i=1;i<n;++i) for(int j=1;j<=m&&B;++j)
		if(c[i][j]=='.'&&c[i+1][j]=='.') --B,c[i][j]='^',c[i+1][j]='v';
	if(B||A) return 0*puts("NO");
	puts("YES");
	for(int i=1;i<=n;++i) puts(c[i]+1);
	return 0;
}