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:
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define MT make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define RT return
#define vv(a,b,c,d) vector<vector<a> >(b,vector<a>(c,d))
#define vvv(a,b,c,d,e) vector<vector<vector<a> > >(b,vv(a,c,d,e))
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;

int N, M, A, B;
char X[1001][1001];

void puta(int x, int y) {
    --A;
    X[x][y] = '<';
    X[x][y + 1] = '>';
}

void putb(int x, int y) {
    --B;
    X[x][y] = '^';
    X[x + 1][y] = 'v';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);
    cin >> N >> M >> A >> B;

    MEM(X, '.');

    if (M % 2)for (int i = 0; i < N - 1 && B>0; i += 2)
        putb(i, 0);
    if (N % 2)for (int j = M % 2; j < M - 1 && A>0; j += 2)
        puta(0, j);

    for (int i = N % 2; i < N - 1; i += 2)for (int j = M - 2; j >= 0; j -= 2) {
        if (A >= 2) {
            for (int k = 0; k < 2 && A>0; ++k)
                puta(i + k, j);
        } else if (B >= 2) {
            for (int k = 0; k < 2 && B>0; ++k)
                putb(i, j + k);
        } else if (A == 1 && B == 1 && N%2 && M%2 && i==N-2&&j==1) {
            puta(i + 1, j - 1);
            putb(i, j + 1);
        } else if (A == 1) {
            puta(i, j);
        } else if (B == 1) {
            putb(i, j);
        }
    }

    if (A == 0 && B == 0) {
        cout << "YES" << endl;
        rep(i, N) {
            rep(j, M)cout << X[i][j];
            cout << endl;
        }
    } else {
        cout << "NO" << endl;
    }
}