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>

using namespace std;

char ans[1000][1001];

void puta(int x, int y, int &a) {
    if (a == 0) return;
    a--;
    ans[x][y] = '<';
    ans[x][y + 1] = '>';
}

void putb(int x, int y, int &b) {
    if (b == 0) return;
    b--;
    ans[x][y] = '^';
    ans[x + 1][y] = 'v';
}

int main() {
    int n, m, a, b, na, nb, s, i, j;
    
    scanf("%d %d %d %d", &n, &m, &a, &b);
    
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            ans[i][j] = '.';
        }
    }
    
    if (n % 2 == 1 && m % 2 == 1) {
        s = (n - 1) * (m - 1);
        
        for (i = 0; i < m - 3; i += 2) puta(n - 1, i, a);
        for (i = 0; i < n - 3; i += 2) putb(i, m - 1, b);
        
        if (a % 2 == 0 && b % 2 == 0 && n > 1 && m > 1) {
            s -= 4;
            puta(n - 3, m - 3, a);
            puta(n - 1, m - 2, a);
            putb(n - 3, m - 1, b);
            putb(n - 2, m - 3, b);
        } else {
            if (m - 3 >= 0) puta(n - 1, m - 3, a);
            if (n - 3 >= 0) putb(n - 3, m - 1, b);
        }
        
        na = a, nb = b;
        
        if (a % 2 == 1) na++;
        if (b % 2 == 1) nb++;
        
        if (s < (na + nb) * 2) {
            puts("NO");
            return 0;
        }
        
        for (i = 0; i + 1 < n; i += 2) {
            for (j = 0; j + 1 < m; j += 2) {
                if (a > 0) {
                    puta(i, j, a);
                    puta(i + 1, j, a);
                } else {
                    putb(i, j, b);
                    putb(i, j + 1, b);
                }
            }
        }
    } else {
        s = n * m;
        
        if (n % 2 == 1) {
            s -= m;
            for (i = 0; i < m; i += 2) puta(n - 1, i, a);
        }
        
        if (m % 2 == 1) {
            s -= n;
            for (i = 0; i < n; i += 2) putb(i, m - 1, b);
        }
        
        na = a, nb = b;
        if (a % 2 == 1) na++;
        if (b % 2 == 1) nb++;
        
        if (s < (na + nb) * 2) {
            puts("NO");
            return 0;
        }
        
        for (i = 0; i + 1 < n; i += 2) {
            for (j = 0; j + 1 < m; j += 2) {
                if (a > 0) {
                    puta(i, j, a);
                    puta(i + 1, j, a);
                } else {
                    putb(i, j, b);
                    putb(i, j + 1, b);
                }
            }
        }
    }
    
    puts("YES");
    for (i = 0; i < n; i++) printf("%s\n", ans[i]);
    
    return 0;
}