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 <bits/stdc++.h>
using namespace std;
const int N=1100;
int n,m,a,b,cnt;
char s[N][N];
int main()
{
	scanf("%d%d%d%d",&n,&m,&a,&b);
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=m;j++) s[i][j]='.';
	}
	cnt=(n/2)*(m/2);
	if (m&1)
	{
		int tmp=min(b,n/2);
		b-=tmp;
		for (int i=1,x=1;i<=tmp;i++,x+=2) s[x][m]='^',s[x+1][m]='v';
	}
	if (n&1)
	{
		int tmp=min(a,m/2);
		a-=tmp;
		for (int i=1,x=1;i<=tmp;i++,x+=2) s[n][x]='<',s[n][x+1]='>';
	}
	if (n&1 && m&1 && a&1 && b&1 && (a+1)/2+(b+1)/2==cnt+1 && n!=1 && m!=1)
	{
		printf("YES\n");
		int x=1,y=1;
		while (a>1)
		{
			s[x][y]='<',s[x][y+1]='>';
			s[x+1][y]='<',s[x+1][y+1]='>';
			a-=2;y+=2;
			if (y>m-(m&1)) y=1,x+=2;
		}
		while (b>1)
		{
			s[x][y]='^',s[x+1][y]='v';
			s[x][y+1]='^',s[x+1][y+1]='v';
			b-=2;y+=2;
			if (y>m-(m&1)) y=1,x+=2;
		}
		s[n][m]='>';s[n][m-1]='<';
		s[n][m-2]='v';s[n-1][m-2]='^';
		s[n-2][m-2]='<';s[n-2][m-1]='>';
		for (int i=1;i<=n;i++)
		{
			for (int j=1;j<=m;j++) printf("%c",s[i][j]);
			printf("\n");
		}
		return 0;
	}
	if ((a+1)/2+(b+1)/2>cnt)
	{
		printf("NO\n");
		return 0;
	}
	printf("YES\n");
	int x=1,y=1;
	while (a>0)
	{
		s[x][y]='<',s[x][y+1]='>';
		if (a!=1) s[x+1][y]='<',s[x+1][y+1]='>';
		a-=2;y+=2;
		if (y>m-(m&1)) y=1,x+=2;
	}
	while (b>0)
	{
		s[x][y]='^',s[x+1][y]='v';
		if (b!=1) s[x][y+1]='^',s[x+1][y+1]='v';
		b-=2;y+=2;
		if (y>m-(m&1)) y=1,x+=2;
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=m;j++) printf("%c",s[i][j]);
		printf("\n");
	}
}