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<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef vector<int> vi;
typedef pair<int,int> pii;
const int N=1000+100;
const db pi=acos(-1.0);
#define lowbit(x) ((x)&(-x))
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u],v=sq[i].to;i;i=sq[i].nxt,v=sq[i].to)
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define maxd 998244353
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

inline ll readll()
{
    ll x=0;int f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

char ans[N][N];

int main()
{
    int n=read(),m=read(),A=read(),B=read();
    int opn=0,opm=0;
    rep(i,1,n) rep(j,1,m) ans[i][j]='.';
    if (n&1)
    {
        for (int i=2;i<=m && A;i+=2)
        {   
            ans[n][i-1]='<';ans[n][i]='>';A--;
        }
        n--;opn=1;
    }
    if (m&1)
    {
        for (int i=2;i<=n && B;i+=2)
        {
            ans[i-1][m]='^';ans[i][m]='v';B--;
        }
        m--;opm=1;
    }
    if ((opn) && (opm) && ((A+B)*2==n*m) && (A&1))
    {
        ans[n-1][m-1]='<';ans[n-1][m]='>';ans[n-1][m+1]='^';
        ans[n][m-1]='^';ans[n][m+1]='v';
        ans[n+1][m-1]='v';ans[n+1][m]='<';ans[n+1][m+1]='>';
        A--;B--;
    }
    //cout << "rst " << A << " " << B << endl;
    for (int i=2;i<=n;i+=2)
    {
        for(int j=2;j<=m;j+=2)
        {
            if (A)
            {
                ans[i-1][j-1]='<';ans[i-1][j]='>';A--;
                if (A) {ans[i][j-1]='<';ans[i][j]='>';A--;}
            }
            else if (B)
            {
                ans[i-1][j-1]='^';ans[i][j-1]='v';B--;
                if (B) {ans[i-1][j]='^';ans[i][j]='v';B--;}
            }
        }
    }
    if (A+B) {puts("NO");return 0;}
    puts("YES");
    n+=opn;m+=opm;
    rep(i,1,n)
    {
        rep(j,1,m) putchar(ans[i][j]);
        puts("");
    }
    return 0;
}