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<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1005;
char map[MAXN][MAXN];
int n,m,a,b,N,M;
void pr()
{
	for(int i=1;i<=N;i++)
	{
		for(int j=1;j<=M;j++)
		{
			printf("%c",map[i][j]==0?'.':map[i][j]);
		}
		printf("\n");
	}
}
int main()
{
	scanf("%d %d %d %d",&n,&m,&a,&b);
	N=n,M=m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			map[i][j]='.';
	if(n%2==1)
	{
		for(int i=1;i<m&&a>0;i+=2)
		{
			map[n][i]='<';
			map[n][i+1]='>';
			a--;
		}
	}
	if(m%2==1)
	{
		for(int j=1;j<n&&b>0;j+=2)
		{
			map[j][m]='^';
			map[j+1][m]='v';
			b--;
		}
	}
	//pr();
	if(n%2==1&&m%2==1&&a%2==1&&b%2==1)
	{
		map[n-2][m-2]='<';
		map[n-2][m-1]='>';
		map[n-2][m   ]='^';
		map[n-1][m-2]='^';
		map[n-1][m   ]='v';
		map[n   ][m-2]='v';
		map[n   ][m-1]='<';
		map[n   ][m   ]='>';
		a--;
		b--;
	}
	for(int i=1;i<=N&&b;i+=2)
	{
		for(int j=1;j<=M&&b;j++)
		{
			if(map[i+1][j]=='.'&&map[i][j]=='.')
			{
				map[i][j]='^';
				map[i+1][j]='v';
				b--;
			}
		}
	}
	for(int i=1;i<=N&&a;i+=2)
	{
		for(int j=1;j<=M&&a;j++)
		{
			if(map[i][j+1]=='.'&&map[i][j]=='.')
			{
				map[i][j]='<';
				map[i][j+1]='>';
				a--;
			}
			if(map[i+1][j+1]=='.'&&map[i+1][j]=='.'&&a)
			{
				map[i+1][j]='<';
				map[i+1][j+1]='>';
				a--;
			}
		}
	}
	//pr();
	if(a||b)
	{
		printf("NO\n");
		return 0;
	}
	printf("YES\n");
	pr();
}