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<cstdlib>
#include<cstring>
#include<iostream>
using namespace std;
int n,m,a,b;
char s[1010][1010];
bool check1(int x,int y)
{
	if(y==m||a==0) return false;
	return s[x][y+1]=='.';
}
bool check2(int x,int y)
{
	if(x==n||b==0) return false;
	if(b==1)
	{
		if(((m%2)&&(y%2==0))||((m%2==0)&&(y%2)))
		{
			s[n-1][m]='^';s[n][m]='v';b--;
			return false;
		}
	}
	return s[x+1][y]=='.';
}
void dfs(int x,int y)
{
	/*printf("x y:%d %d\n",x,y);
	for(int i=1;i<=n;i++) printf("%s\n",s[i]+1);
	printf("-----------------\n");*/
	if(x==n+1) return;
	if(s[x][y]=='.')
	{
		if(check2(x,y)) s[x][y]='^',s[x+1][y]='v',b--;
		else if(check1(x,y)) s[x][y]='<',s[x][y+1]='>',a--;
	}
	if(y==m) dfs(x+1,1);
	else dfs(x,y+1);
}
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]='.';
	if(m%2)
	{
		for(int i=1;i<=n;i+=2)
		{
			if(i==n-1)
			{
				if(b%2==1&&check2(i,1)) s[i][1]='^',s[i+1][1]='v',b--;
			}
			if(check2(i,1)) s[i][1]='^',s[i+1][1]='v',b--;
		}
		//for(int j=1;j<=m;j++) if(check1(n,j)) s[n][j]='<',s[n][j+1]='>',a--;
	}
	dfs(1,1);
	//for(int i=1;i<=n;i++) printf("%s\n",s[i]+1);
	if(a!=0||b!=0) printf("NO\n");
	else
	{
		printf("YES\n");
		for(int i=1;i<=n;i++) printf("%s\n",s[i]+1);
	}
}