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<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define SF scanf
#define PF printf
#define MAXN 1010
using namespace std;
int ans[MAXN][MAXN];
char w[10]={'.','<','>','^','v'};
int n,m;
int ablx,ably;
bool work(int a,int b,int flag){
    if(n%2==1)
        ablx=m/2;
    if(m%2==1)
        ably=n/2;
    if(flag==1){
        ablx--;
        ably--;
    }
    for(int i=1;i+1<=n;i+=2)
        for(int j=1;j+1<=m;j+=2){
            if(ans[i][j]||ans[i][j+1]||ans[i+1][j]||ans[i+1][j+1])
                continue;
            if(a>ablx){
                ans[i][j]=1;
                ans[i][j+1]=2;
                a--;
                if(a>0){
                    ans[i+1][j]=1;
                    ans[i+1][j+1]=2;
                    a--;
                }
            }
            else if(b>ably){
                ans[i][j]=3;
                ans[i+1][j]=4;
                b--;
                if(b>0){
                    ans[i][j+1]=3;
                    ans[i+1][j+1]=4;
                    b--;
                }
            }
        }
    if(n%2==1){
        for(int j=1;j+1<=m;j+=2)
            if(a>0&&ans[n][j]==0&&ans[n][j+1]==0){
                ans[n][j]=1;
                ans[n][j+1]=2;
                a--;
            }
    }
    if(m%2==1){
        for(int i=1;i+1<=n;i+=2)
            if(b>0&&ans[i][m]==0&&ans[i+1][m]==0){
                ans[i][m]=3;
                ans[i+1][m]=4;
                b--;
            }
    }
    return a||b;
}
void print(){
    PF("YES\n");
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)
            PF("%c",w[ans[i][j]]);
        PF("\n");
    }
}
int main(){
    int a,b;
    SF("%d%d%d%d",&n,&m,&a,&b);
    if(work(a,b,0)==0){
        print();
    }
    else if(n%2==1&&m%2==1&&n>=3&&m>=3&&a>=2&&b>=2){
            memset(ans,0,sizeof ans);
            ans[n][m]=4;
            ans[n-1][m]=3;
            ans[n-2][m]=2;
            ans[n-2][m-1]=1;
            ans[n-2][m-2]=3;
            ans[n-1][m-2]=4;
            ans[n][m-2]=1;
            ans[n][m-1]=2;
            if(work(a-2,b-2,1)==0)
                print();
            else
                PF("NO");
        }
    else{
        PF("NO");
    }
}