#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 1024;
int N,M;
int A,B;
string ans[MC];

int main(){
	cin >> N >> M >> A >> B;
	int x = 0 , y = 0 , z = (N/2) * (M/2);
	if(N&1) x = min(M/2,A);
	if(M&1) y = min(N/2,B);
	int p = (A-x+1) / 2;
	int q = (B-y+1) / 2;
	bool f = 0;
	if(p+q>z){
		if((A-x)%2==1 && (B-y)%2==1 && N%2==1 && M%2==1 && p+q==z+1){
			A -= 2;
			B -= 2;
			--x;
			--y;
			f = 1;
		} else {
			return cout << "NO" << endl , 0;
		}
	}
	cout << "YES" << endl;
	repp(i,0,N) repp(j,0,M) ans[i] += ".";
	A -= x;
	B -= y;
	for(int i = 0 ; i+1 < N ; i += 2) for(int j = 0 ; j+1 < M ; j += 2){
		
		if(A){
			ans[i][j] = '<'; ans[i][j+1] = '>';
			--A;
			if(A){
				ans[i+1][j] = '<'; ans[i+1][j+1] = '>';
				--A;
			}
		} else if(B){
			ans[i][j] = '^'; ans[i+1][j] = 'v';
			--B;
			if(B){
				ans[i][j+1] = '^'; ans[i+1][j+1] = 'v';
				--B;
			}
		}
	}
	repp(i,0,x){
		ans[N-1][i*2] = '<'; ans[N-1][i*2+1] = '>';
	}
	repp(i,0,y){
		ans[i*2][M-1] = '^'; ans[i*2+1][M-1] = 'v';
	}
	if(f){
		ans[N-3][M-3] = '<'; ans[N-3][M-2] = '>';
		ans[N-1][M-2] = '<'; ans[N-1][M-1] = '>';
		ans[N-2][M-3] = '^'; ans[N-1][M-3] = 'v';
		ans[N-3][M-1] = '^'; ans[N-2][M-1] = 'v';
	}
	repp(i,0,N) cout << ans[i] << endl;
	return 0;
}

### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 1024;
int N,M;
int A,B;
string ans[MC];

int main(){
	cin >> N >> M >> A >> B;
	int x = 0 , y = 0 , z = (N/2) * (M/2);
	if(N&1) x = min(M/2,A);
	if(M&1) y = min(N/2,B);
	int p = (A-x+1) / 2;
	int q = (B-y+1) / 2;
	bool f = 0;
	if(p+q>z){
		if((A-x)%2==1 && (B-y)%2==1 && N%2==1 && M%2==1 && p+q==z+1){
			A -= 2;
			B -= 2;
			--x;
			--y;
			f = 1;
		} else {
			return cout << "NO" << endl , 0;
		}
	}
	cout << "YES" << endl;
	repp(i,0,N) repp(j,0,M) ans[i] += ".";
	A -= x;
	B -= y;
	for(int i = 0 ; i+1 < N ; i += 2) for(int j = 0 ; j+1 < M ; j += 2){
		
		if(A){
			ans[i][j] = '<'; ans[i][j+1] = '>';
			--A;
			if(A){
				ans[i+1][j] = '<'; ans[i+1][j+1] = '>';
				--A;
			}
		} else if(B){
			ans[i][j] = '^'; ans[i+1][j] = 'v';
			--B;
			if(B){
				ans[i][j+1] = '^'; ans[i+1][j+1] = 'v';
				--B;
			}
		}
	}
	repp(i,0,x){
		ans[N-1][i*2] = '<'; ans[N-1][i*2+1] = '>';
	}
	repp(i,0,y){
		ans[i*2][M-1] = '^'; ans[i*2+1][M-1] = 'v';
	}
	if(f){
		ans[N-3][M-3] = '<'; ans[N-3][M-2] = '>';
		ans[N-1][M-2] = '<'; ans[N-1][M-1] = '>';
		ans[N-2][M-3] = '^'; ans[N-1][M-3] = 'v';
		ans[N-3][M-1] = '^'; ans[N-2][M-1] = 'v';
	}
	repp(i,0,N) cout << ans[i] << endl;
	return 0;
}

```