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;
char a[1010][1010];
int main(){
//	freopen("C.in","r",stdin);
	int n,m,A,B;
	int cntA,cntB;
	scanf("%d%d%d%d",&n,&m,&A,&B);
	if(n&1){
		if(A>m/2)
			cntA=m/2,A-=m/2;
		else
			cntA=A,A=0;
	}
	else
		cntA=0;

	if(m&1){
		if(B>n/2)
			cntB=n/2,B-=n/2;
		else
			cntB=B,B=0;
	}
	else
		cntB=0;
	if((A&1)&&(B&1)&&(n&1)&&(m&1)&&A/2+B/2<(n/2)*(m/2)){
		A--,B--;
		for(int i=1;i<n;i+=2){
			for(int j=1;j<m;j+=2){
				if(i+2>=n&&j+2>=m) break;
				if(A>0){
					if(A>=2){
						a[i][j]='<',a[i][j+1]='>';
						a[i+1][j]='<',a[i+1][j+1]='>';
						A-=2;
					}
					else{
						a[i][j]='<',a[i][j+1]='>';
						a[i+1][j]='.',a[i+1][j+1]='.';
						A--;
					}
				}
				else if(B>0){
					if(B>=2){
						a[i][j]='^',a[i+1][j]='v';
						a[i][j+1]='^',a[i+1][j+1]='v';
						B-=2;
					}
					else{
						a[i][j]='^',a[i+1][j]='v';
						a[i][j+1]='.',a[i+1][j+1]='.';
						B--;
					}
				}
				else{
					a[i][j]='.',a[i+1][j]='.';
					a[i][j+1]='.',a[i+1][j+1]='.';
				}
			}
		}
		for(int i=1;i<=(cntA-1)*2;i+=2)
			a[n][i]='<',a[n][i+1]='>';
		for(int i=1;i<=(cntB-1)*2;i+=2)
			a[i][m]='^',a[i+1][m]='v';
		a[n][m-1]='<',a[n][m]='>';
		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';
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
				if(a[i][j]==0)
					a[i][j]='.';
		printf("YES\n");
		for(int i=1;i<=n;i++){
			for(int j=1;j<=m;j++)
				printf("%c",a[i][j]);
			printf("\n");
		}
		return 0;
	}
	if((A+1)/2+(B+1)/2>(n/2)*(m/2)){
		printf("NO\n");
		return 0;
	}
	for(int i=1;i<n;i+=2){
		for(int j=1;j<m;j+=2){
			if(A>0){
				if(A>=2){
					a[i][j]='<',a[i][j+1]='>';
					a[i+1][j]='<',a[i+1][j+1]='>';
					A-=2;
				}
				else{
					a[i][j]='<',a[i][j+1]='>';
					a[i+1][j]='.',a[i+1][j+1]='.';
					A--;
				}
			}
			else if(B>0){
				if(B>=2){
					a[i][j]='^',a[i+1][j]='v';
					a[i][j+1]='^',a[i+1][j+1]='v';
					B-=2;
				}
				else{
					a[i][j]='^',a[i+1][j]='v';
					a[i][j+1]='.',a[i+1][j+1]='.';
					B--;
				}
			}
			else{
				a[i][j]='.',a[i+1][j]='.';
				a[i][j+1]='.',a[i+1][j+1]='.';
			}
		}
	}
	for(int i=1;i<=cntA*2;i+=2)
		a[n][i]='<',a[n][i+1]='>';
	for(int i=1;i<=cntB*2;i+=2)
		a[i][m]='^',a[i+1][m]='v';
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(a[i][j]==0)
				a[i][j]='.';
	printf("YES\n");
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++)
			printf("%c",a[i][j]);
		printf("\n");
	}
	return 0;
}