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<cstdio>
#include<iostream>
#define maxn 1010
using namespace std;
int n,m,i,j,x,y,s,sa,sb;
char a[maxn][maxn];
int main()
{
	scanf("%d%d%d%d",&n,&m,&x,&y);
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
			a[i][j]='.';
	if(n*m&1)
	{
		s=n*m>>1;
		sa=n*(m>>1);
		sb=m*(n>>1);
		if(x+y>s||x>sa||y>sb)
		{
			puts("NO");
			return 0;
		}
		if((sa-x)%2==1&&(y==s-x)&&x>=2&&y>=2&&n>=3&&m>=3)
		{
			a[n-2][m-2]='<';a[n-2][m-1]='>';
			a[n-1][m-2]='^';a[n][m-2]='v';
			a[n-2][m]='^';a[n-1][m]='v';
			a[n][m-1]='<';a[n][m]='>';
			x-=2;y-=2;
			for(i=1;i<m-2;i+=2)
				if(x)
				{
					x--;
					a[n][i]='<';
					a[n][i+1]='>';
				}
			for(i=1;i<n-2;i+=2)
				if(y)
				{
					y--;
					a[i][m]='^';
					a[i+1][m]='v';
				}
			for(i=1;i<n;i+=2)
				for(j=1;j<m;j+=2)
				{
					if(i==n-2&&j==m-2)break;
					if(x>1)
					{
						a[i][j]='<';
						a[i][j+1]='>';
						a[i+1][j]='<';
						a[i+1][j+1]='>';
						x-=2;
					}else
					if(x==1)
					{
						a[i][j]='<';
						a[i][j+1]='>';
						//i--;
						x--;
					}else
					if(y>1)
					{
						a[i][j]='^';
						a[i][j+1]='^';
						a[i+1][j]='v';
						a[i+1][j+1]='v';
						y-=2;
					}else
					if(y==1)
					{
						a[i][j]='^';
						a[i+1][j]='v';
						//j--;
						y--;
					}
				}
					
			puts("YES");
			for(i=1;i<=n;i++)
			{
				for(j=1;j<=m;j++)
					putchar(a[i][j]);
				putchar('\n');
			}
			return 0;
		}
	}
	if(n&1)
		for(i=1;i+3<=m;i+=4)
		{
			if(x<2)break;
			a[n][i]='<';
			a[n][i+1]='>';
			a[n][i+2]='<';
			a[n][i+3]='>';
			x-=2;
		}
	if(n%2==1&&x%2==1&&i<m)
	{
		a[n][i]='<';
		a[n][i+1]='>';
		x--;
	}
	if(m&1)
		for(i=1;i+3<=n;i+=4)
		{
			if(y<2)break;
			a[i][m]='^';
			a[i+1][m]='v';
			a[i+2][m]='^';
			a[i+3][m]='v';
			y-=2;
		}
	if(m%2==1&&y%2==1&&i<n)
	{
		a[i][m]='^';
		a[i+1][m]='v';
		y--;
	}
	for(i=1;i<n;i+=2)
		for(j=1;j<m;j+=2)
		{
			if(x>1)
			{
				a[i][j]='<';
				a[i][j+1]='>';
				a[i+1][j]='<';
				a[i+1][j+1]='>';
				x-=2;
			}else
			if(x==1)
			{
				a[i][j]='<';
				a[i][j+1]='>';
				//i--;
				x--;
			}else
			if(y>1)
			{
				a[i][j]='^';
				a[i][j+1]='^';
				a[i+1][j]='v';
				a[i+1][j+1]='v';
				y-=2;
			}else
			if(y==1)
			{
				a[i][j]='^';
				a[i+1][j]='v';
				//j--;
				y--;
			}
		}
	if(x>0||y>0)
	{
		puts("NO");
		return 0;
	}
	puts("YES");
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=m;j++)
			putchar(a[i][j]);
		putchar('\n');
	}
}