Problem:
Score : 200 points

Problem Statement

Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.

The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.

One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.

If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.

X < Z \leq Y

x_1, x_2, ..., x_N < Z

y_1, y_2, ..., y_M \geq Z

Determine if war will break out.

Constraints

All values in input are integers.

1 \leq N, M \leq 100

-100 \leq X < Y \leq 100

-100 \leq x_i, y_i \leq 100

x_1, x_2, ..., x_N \neq X

x_i are all different.

y_1, y_2, ..., y_M \neq Y

y_i are all different.

Input

Input is given from Standard Input in the following format:

N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M

Output

If war will break out, print War; otherwise, print No War.

Sample Input 1

3 2 10 20
8 15 13
16 22

Sample Output 1

No War

The choice Z = 16 satisfies all of the three conditions as follows, thus they will come to an agreement.

X = 10 < 16 \leq 20 = Y

8, 15, 13 < 16

16, 22 \geq 16

Sample Input 2

4 2 -48 -1
-20 -35 -91 -23
-22 66

Sample Output 2

War

Sample Input 3

5 3 6 8
-10 3 1 5 -100
100 6 14

Sample Output 3

War

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
    int n,m,x,y,i,j1=0,k=0,l=0;
    scanf("%d%d%d%d",&n,&m,&x,&y);
    int xx[n],yy[m];
    for(i=0; i<n; i++)
    {
        scanf("%d",&xx[i]);
    }
    for(i=0; i<m; i++)
    {
        scanf("%d",&yy[i]);
    }
    for(i=x+1; i<=y; i++)
    {
        j1++;
    }
    int z[j1],j;
    for(i=x+1; i<=y; i++)
    {
        z[k]=i;
        k++;
    }
    k=0;
    for(i=0; i<j1; i++)
    {
        if(k==0)
        {
            for(j=0; j<n; j++)
            {
                if(xx[j]<z[i])
                {
                    k=z[i];
                }
                else
                {
                    k=0;
                    z[i]=200;
                    break;
                }
            }
        }
        else
        {
            break;
        }

    }
    for(i=0; i<j1; i++)
    {
        for(j=0; j<m; j++)
        {
            if(yy[j]>=z[i] && z[i]!=200)
            {
                l++;
            }
            else
            {
                break;
            }
        }
    }
    if(l==m || l==(j1*m))
    {
        printf("No War\n");
    }
    else
    {
        printf("War\n");
    }
    return 0;
}