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 <stdio.h>
#include <algorithm>
using namespace std;
#define N (1010)
int n, m, a, b;
int ans[N][N];
char s[N][N];
int cnt;
void print(){
    for (int i = 1; i <= n; ++i){
        for (int j = 1; j <= m; ++j){
            if (!ans[i][j]){
                s[i-1][j-1] = '.';
            }
            else{
                if (ans[i+1][j] == ans[i][j]){
                    s[i-1][j-1] = '^';
                    s[i][j-1] = 'v';
                }
                else if (ans[i][j+1] == ans[i][j]){
                    s[i-1][j-1] = '<';
                    s[i-1][j] = '>';
                }
                else {
                    s[i][j] = '?';
                }
            }
        }
        s[i-1][m] = 0;
    }
    for (int i = 0; i < n; ++i){
        printf("%s\n", s[i]);
    }
}
#define ver(x,y) ans[x][y]=ans[x+1][y]=++cnt, --b;
#define hor(x,y) ans[x][y]=ans[x][y+1]=++cnt, --a;
int main(){
    scanf("%d%d%d%d", &n, &m, &a, &b);
    if (m & 1)
    for (int i = 1; i+1 <= n && b; i+=2){
        ver(i,m);
    }
    if (n & 1)
    for (int i = 1; i+1 <= m && a; i+=2){
        hor(n,i);
    }
    int nn = n/2*2, mm = m/2*2;
    for (int i = 1; i+1 <= nn; i+=2){
        for (int j = 1; j+1 <= mm; j+=2){
            if (a >= 2) {
                hor(i,j);
                hor(i+1,j);
                continue;
            }
            if (b >= 2){
                ver(i,j);
                ver(i,j+1);
                continue;
            }
            if (a){
                hor(i,j);
                continue;
            }
            if (b){
                ver(i,j);
            }
        }
    }
    if ((n&m&1) && b==1 && !a){
        if (n>1 && m>1 && ans[n-2][m-2]==ans[n-2][m-1]){
            hor(n-2,m-2);
            hor(n,m-1);
            ver(n-1,m-2);
            ver(n-2,m);
            a = b = 0;
        }
    }
    if (a || b){
        printf("NO\n");
    }
    else {
        printf("YES\n");
        print();
    }
}