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>
#define rg register
const int N=3000;

using namespace std;

char ans[N][N];
int n,m,a,b,fl=1;

namespace __{
	void solve(){
		//freopen("c.in","r",stdin);
		//freopen("c.out","w",stdout);
		scanf("%d%d%d%d",&n,&m,&a,&b);
		int nn=n,mm=m;
		for(int i=0; i<n; ++i)
			for(int j=0; j<m; ++j)
				ans[i][j]='.';

		if(n&1){			
			for(int i=0; i+1<m; i+=2){
				if(a){
					ans[n-1][i]='<';
					ans[n-1][i+1]='>';--a;
				}
			} --n;
		} else fl=0;
		if(m&1){
			for(int i=0; i+1<n; i+=2){
				if(b){
					ans[i][m-1]='^';
					ans[i+1][m-1]='v';--b;
				}
			} --m;
		} else fl=0;


		if((fl) && (a+(a&1)+b+(b&1)>(n*m)/2)){
		 	++n; ++m;
			ans[n-1][m-3]=ans[n-3][m-2]='<';
			ans[n-1][m-2]=ans[n-3][m-1]='>';
			ans[n-3][m-3]=ans[n-2][m-1]='^';
			ans[n-2][m-3]=ans[n-1][m-1]='v';
			--n;--m;a-=1;b-=1;
		}

		for(int i=0; i<n; i+=2)	
			for(int j=0; j<m; j+=2)if(ans[i][j]=='.'){
				if(a){
					a--;
					ans[i][j]='<';
					ans[i][j+1]='>';
					if(a){
						ans[i+1][j]='<';
						ans[i+1][j+1]='>';
						--a;
					}
				}
				else if(b){
					b--;
					ans[i][j]='^';
					ans[i+1][j]='v';
					if(b){
						ans[i][j+1]='^';
						ans[i+1][j+1]='v';
						--b;
					}
				}
			}
		if(!(a|b)){
			puts("YES");
			for(rg int i=0; i<nn; ++i) puts(ans[i]);
		}
		else{
			puts("NO");
		}
	}
}
int main(){
	__::solve();
}