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;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define enum(i,x,y) for(int i=(x);i<=(y);++i)
#define try(i,x,y) for(int i=(x);i>=(y);--i)
template <typename TYPE> inline void chkmax(TYPE &x,TYPE y){x<y?x=y:0;}
template <typename TYPE> inline void chkmin(TYPE &x,TYPE y){y<x?x=y:0;}
template <typename TYPE> void readint(TYPE &x)
{
    x=0;int f=1;char c;
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
    for(;isdigit(c);c=getchar())x=x*10+c-'0';
    x*=f;
}
const int MAXN=1005;

int n,m,a,b;
char s[MAXN][MAXN];
void chara(int i,int j)
{
	if(a>0)
	{
		a-=2;
		s[i][j]='<',s[i][j+1]='>';
		s[i+1][j]=(a>=0?'<':'.'),s[i+1][j+1]=(a>=0?'>':'.');
		if(a<0)a=0;
	}
	else if(b>0)
	{
		b-=2;
		s[i][j]='^',s[i][j+1]=(b>=0?'^':'.');
		s[i+1][j]='v',s[i+1][j+1]=(b>=0?'v':'.');
		if(b<0)b=0;
	}
	else s[i][j]=s[i][j+1]=s[i+1][j]=s[i+1][j+1]='.';
}
bool work(int n,int m)
{
	if(2*(a+b)==n*m && (a&1))return 0;
	enum(i,1,n/2)
		enum(j,1,m/2)
			chara(i*2-1,j*2-1);
	if(a!=0 || b!=0)return 0;
	return 1;
}
void print()
{
	printf("YES\n");
	enum(i,1,n)
	{
		enum(j,1,m)
			putchar(s[i][j]);
		putchar('\n');
	}
}
void solve()
{
	s[n-2][m-2]='<';s[n-2][m-1]='>';s[n-2][m]='^';
	s[n-1][m-2]='^';s[n-1][m-1]='.';s[n-1][m]='v';
	s[n][m-2]='v';s[n][m-1]='<';s[n][m]='>';
	a-=2;b-=2;
	enum(i,1,(m-3)/2){s[n][i*2-1]='<';s[n][i*2]='>';--a;}
	enum(i,1,(n-3)/2){s[i*2-1][m]='^';s[i*2][m]='v';--b;}
	enum(i,1,(n-1)/2)
		enum(j,1,(m-1)/2)
			if(i!=(n-1)/2 || j!=(m-1)/2)chara(i*2-1,j*2-1);
	if(a!=0 || b!=0){printf("NO\n");exit(0);}
}

int main()
{
	readint(n);readint(m);readint(a);readint(b);
	int ta=a,tb=b;
	if((a+b)*2>n*m)return 0*printf("NO\n");
	s[n][m]='.';
	if(n&1)
	{
		enum(i,1,m/2)
			if(a>0){--a;s[n][i*2-1]='<';s[n][i*2]='>';}
			else s[n][i*2-1]=s[n][i*2]='.';
	}
	if(m&1)
	{
		enum(i,1,n/2)
			if(b>0){--b;s[i*2-1][m]='^';s[i*2][m]='v';}
			else s[i*2-1][m]=s[i*2][m]='.';
	}
	if(work(n/2*2,m/2*2))print();
	else if((n&1) && (m&1) && n>1 && m>1){a=ta;b=tb;solve();print();}
	else printf("NO\n");
    return 0;
}