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;
int n, m, a, b;
char o[1005][1005];

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++) {
			o[i][j] = '.';
		}
	}
	if(n%2 == 1) {
		for(int i=1;i+1<=m;i+=2) {
			if(!a) break;
			a--;
			o[1][i] = '<';
			o[1][i+1] = '>';
		}
	}
	if(m%2 == 1) {
		for(int i=1;i+1<=n;i+=2) {
			if(!b) break;
			b--;
			o[i][m] = '^';
			o[i+1][m] = 'v';
		}
	}
	for(int i=1+n%2;i+1<=n;i+=2) {
		for(int j=1;j+1<=m/2*2;) {
			if(b >= 2) {
				o[i][j] = '^';
				o[i+1][j] = 'v';
				o[i][j+1] = '^';
				o[i+1][j+1] = 'v';
				b -= 2; j += 2;
			}
			else if(a >= 2) {
				o[i][j] = '<';
				o[i][j+1] = '>';
				o[i+1][j] = '<';
				o[i+1][j+1] = '>';
				a -= 2; j += 2;
			}
			else if(b) {
				o[i][j] = '^';
				o[i+1][j] = 'v';
				b--; j++;
			}
			else if(a) {
				o[i][j] = '<';
				o[i][j+1] = '>';
				a--; j += 2;
			}
			else j += 2;
		}
	}
	if(a && o[n][m] == '.' && o[n][m-1] == '.') {
		a--;
		o[n][m-1] = '<';
		o[n][m] = '>';
	}
	if(a || b) puts("NO");
	else {
		puts("YES");
		for(int i=1;i<=n;i++) printf("%s\n",o[i]+1);
	}
}