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<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF 1e16
#define mod 1000000007

int A,B,N,M;
string res[1001];

void putA(ll i,ll j){
  res[i][j]='<'; res[i][j+1]='>'; A--;
}
void putB(ll i,ll j){
  res[i][j]='^'; res[i+1][j]='v'; B--;
}


int main(){
  cin>>N>>M>>A>>B;
  int area=N*M,tot=2*(A+B);
  if(area<tot){
    cout<<"NO"<<endl;
    return 0;
  }
  rep(i,N)rep(j,M)res[i]+=".";
  if(N%2==0&&M%2==0){
    for(ll i=0;i<N;i+=2)for(ll j=0;j<M;j+=2){
      if(A>0){
        putA(i,j);
        if(A>0)putA(i+1,j);
      }else if(B>0){
        putB(i,j);
        if(B>0)putB(i,j+1);
      }
    }
  }else if(M%2==0){
    for(ll j=0;j<M;j+=2){
      if(A>0){
        putA(N-1,j);
      }
    }
    for(ll i=0;i<N-1;i+=2)for(ll j=0;j<M;j+=2){
      if(A>0){
        putA(i,j);
        if(A>0)putA(i+1,j);
      }else if(B>0){
        putB(i,j);
        if(B>0)putB(i,j+1);
      }
    }
  }else if(N%2==0){
    for(ll i=0;i<N;i+=2){
      if(B>0){
        putB(i,M-1);
      }
    }
    for(ll i=0;i<N;i+=2)for(ll j=0;j<M-1;j+=2){
      if(A>0){
        putA(i,j);
        if(A>0)putA(i+1,j);
      }else if(B>0){
        putB(i,j);
        if(B>0)putB(i,j+1);
      }
    }
  }else{
    ll preA=A,preB=B;
    for(ll j=0;j<M-1;j+=2){
      if(A>0){
        putA(N-1,j);
      }
    }
    for(ll i=0;i<N-1;i+=2){
      if(B>0){
        putB(i,M-1);
      }
    }
    for(ll i=0;i<N-1;i+=2)for(ll j=0;j<M-1;j+=2){
      if(A>0){
        putA(i,j);
        if(A>0)putA(i+1,j);
      }else if(B>0){
        putB(i,j);
        if(B>0)putB(i,j+1);
      }
    }

    if(A>0||B>0){
      rep(i,N)res[i]="";
      rep(i,N)rep(j,M)res[i]+=".";

      A=preA; B=preB;
      //assert(A>=2&&B>=2);
      if(A>0)putA(N-3,M-3);
      if(B>0)putB(N-3,M-1);
      if(A>0)putA(N-1,M-2);
      if(B>0)putB(N-2,M-3);
      for(ll j=0;j<M-3;j+=2){
        if(A>0){
          putA(N-1,j);
        }
      }
      for(ll i=0;i<N-3;i+=2){
        if(B>0){
          putB(i,M-1);
        }
      }
      for(ll i=0;i<N-1;i+=2)for(ll j=0;j<M-1;j+=2){
        if(i==N-3&&j==M-3)continue;

        if(A>0){
          putA(i,j);
          if(A>0)putA(i+1,j);
        }else if(B>0){
          putB(i,j);
          if(B>0)putB(i,j+1);
        }
      }
    }
  }
  if(A>0||B>0){
    cout<<"NO"<<endl;
    return 0;
  }
  cout<<"YES"<<endl;
  rep(i,N){
    cout<<res[i]<<endl;
  }
  return 0;
}