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 <iostream>
#include <cstdio>
#define N 1005
using namespace std;

int n, m, a, b;
char c[N][N];
int main()
{
	int i, j;
	cin >> n >> m >> a >> b;
	for (i = 0; i < n; i++) for (j = 0; j < m; j++) c[i][j] = '.';
	if (m % 2) for (i = 0; i + 1 < n; i += 2) if (b > 0) b--, c[i][m - 1] = '^', c[i + 1][m - 1] = 'v';
	if (n % 2 && n > 2 && m > 2 && b % 2) b--, c[n - 2][m - 3] = '^', c[n - 1][m - 3] = 'v';
	for (i = 0; i < n; i++) {
		for (j = 0; j < m; j++) {
			if (b > 0) {
				if (c[i][j] == '.' && c[i + 1][j] == '.') b--, c[i][j] = '^', c[i + 1][j] = 'v';
			}
		}
	}
	for (i = 0; i < n; i++) {
		for (j = 0; j < m; j++) {
			if (a > 0) {
				if (c[i][j] == '.' && c[i][j + 1] == '.') a--, c[i][j] = '<', c[i][j + 1] = '>';
			}
		}
	}
	if (a > 0 || b > 0) {cout << "NO"; return 0;}
	cout << "YES\n";
	for (i = 0; i < n; i++) printf("%s\n", c[i]);
    return 0;
}