#include <algorithm>
#include <cctype>
#include <cstring>
#include <cstdio>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e3+5;
int n, m, a, b; char s[N][N];
int main()
{
    n = read(), m = read(), a = read(), b = read();
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
            s[i][j] = '.';
    if(n&1) for(int i = 1; i < m&&a; i += 2) --a, s[n][i] = '<', s[n][i+1] = '>';
    if(m&1) for(int i = 1; i < n&&b; i += 2) --b, s[i][m] = '^', s[i+1][m] = 'v';
    for(int i = 1; i < n; i += 2)
        for(int j = 1; j < m; j += 2)
            if(a >= 2) a -= 2, s[i][j] = s[i+1][j] = '<', s[i][j+1] = s[i+1][j+1] = '>';
            else if(b >= 2) b -= 2, s[i][j] = s[i][j+1] = '^', s[i+1][j] = s[i+1][j+1] = 'v';
            else if(a&&b&&i == n-2&&j == m-2) 
                --a, --b,
                s[i][j] = '<', s[i][j+1] = '>', s[i][j+2] = '^', 
                s[i+1][j] = '^', s[i+1][j+2] = 'v',
                s[i+2][j] = 'v', s[i+2][j+1] = '<', s[i+2][j+2] = '>';
            else if(a)  --a, s[i][j] = '<', s[i][j+1] = '>';
            else if(b) --b, s[i][j] = '^', s[i+1][j] = 'v';
    if(a||b) return puts("NO"), 0;
    puts("YES"); for(int i = 1; i <= n; ++i) puts(s[i]+1);
    return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <algorithm>
#include <cctype>
#include <cstring>
#include <cstdio>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e3+5;
int n, m, a, b; char s[N][N];
int main()
{
    n = read(), m = read(), a = read(), b = read();
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
            s[i][j] = '.';
    if(n&1) for(int i = 1; i < m&&a; i += 2) --a, s[n][i] = '<', s[n][i+1] = '>';
    if(m&1) for(int i = 1; i < n&&b; i += 2) --b, s[i][m] = '^', s[i+1][m] = 'v';
    for(int i = 1; i < n; i += 2)
        for(int j = 1; j < m; j += 2)
            if(a >= 2) a -= 2, s[i][j] = s[i+1][j] = '<', s[i][j+1] = s[i+1][j+1] = '>';
            else if(b >= 2) b -= 2, s[i][j] = s[i][j+1] = '^', s[i+1][j] = s[i+1][j+1] = 'v';
            else if(a&&b&&i == n-2&&j == m-2) 
                --a, --b,
                s[i][j] = '<', s[i][j+1] = '>', s[i][j+2] = '^', 
                s[i+1][j] = '^', s[i+1][j+2] = 'v',
                s[i+2][j] = 'v', s[i+2][j+1] = '<', s[i+2][j+2] = '>';
            else if(a)  --a, s[i][j] = '<', s[i][j+1] = '>';
            else if(b) --b, s[i][j] = '^', s[i+1][j] = 'v';
    if(a||b) return puts("NO"), 0;
    puts("YES"); for(int i = 1; i <= n; ++i) puts(s[i]+1);
    return 0;
}
```