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 <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()
#define FINISH cout<<"NO"<<endl;return 0
using ll = long long;
const ll MOD = 1000000007;

char ans[1010][1010];

void tile(int& A, int& B, int N, int M) {
	for (int i = 0;i < N;i += 2) {
		for (int j = 0;j < M;j += 2) {
			if (A > 1) {
				ans[i][j] = '<';
				ans[i][j + 1] = '>';
				ans[i + 1][j] = '<';
				ans[i + 1][j + 1] = '>';
				A -= 2;
			}
			else if (B > 1) {
				ans[i][j] = '^';
				ans[i + 1][j] = 'v';
				ans[i][j + 1] = '^';
				ans[i + 1][j + 1] = 'v';
				B -= 2;
			}
			else if (A == 1) {
				ans[i][j] = '<';
				ans[i][j + 1] = '>';
				A = 0;
			}
			else if (B == 1) {
				ans[i][j] = '^';
				ans[i + 1][j] = 'v';
				B = 0;
			}
		}
	}
}

void say(int N, int M) {
	cout << "YES\n";
	for (int i = 0;i < N;i++) {
		for (int j = 0;j < M;j++)
			cout << ans[i][j];
		cout << "\n";
	}
}

int main() {
	int N, M, A, B;
	cin >> N >> M >> A >> B;
	int T = N * M;
	for (int i = 0;i < N;i++)for (int j = 0;j < M;j++)ans[i][j] = '.';

	if (A + B > T / 2) {
		FINISH;
	}

	if (T & 1) {
		int MA = N * (M / 2);
		int MB = M * (N / 2);
		if (MA < A) {
			FINISH;
		}
		if (MB < B) {
			FINISH;
		}

		//すべてをMAに使う
		for (int i = 0;i < M - 1;i += 2) {
			if (A > 0) {
				ans[N - 1][i] = '<';
				ans[N - 1][i + 1] = '>';
				A--;
			}
		}
		for (int i = 0;i < N - 1;i += 2) {
			if (B > 0) {
				ans[i][M - 1] = '^';
				ans[i + 1][M - 1] = 'v';
				B--;
			}
		}
		tile(A, B, N - 1, M - 1);
		if (B == 1) {
			ans[N - 2][M - 3] = '^';
			ans[N - 1][M - 3] = 'v';
			ans[N - 1][M - 2] = '<';
			ans[N - 1][M - 1] = '>';
			B--;
		}
		say(N, M);
	}
	else {
		T /= 2;
		if (N & 1) {
			for (int i = 0;i < M;i += 2) {
				if (A > 0) {
					ans[N - 1][i] = '<';
					ans[N - 1][i + 1] = '>';
					A--;
				}
			}
			tile(A, B, N - 1, M);
			
			if (A > 0 || B > 0) {
				FINISH;
			}
			say(N, M);
		}
		else if (M & 1) {
			for (int i = 0;i < N;i += 2) {
				if (B > 0) {
					ans[i][M - 1] = '^';
					ans[i + 1][M - 1] = 'v';
					B--;
				}
			}

			tile(A, B, N, M - 1);

			if (A > 0 || B > 0) {
				FINISH;
			}
			say(N, M);
		}
		else {
			tile(A, B, N, M);
			if (A > 0 || B > 0) {
				FINISH;
			}
			say(N, M);
		}
	}
}