Ayoub had an array a of integers of size n and this array had two interesting properties: 

  * All the integers in the array were between l and r (inclusive). 
  * The sum of all the elements was divisible by 3. 



Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array. 

Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.

Input

The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.

Output

Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.

Examples

Input


2 1 3


Output


3


Input


3 2 2


Output


1


Input


9 9 99


Output


711426616

Note

In the first example, the possible arrays are : [1,2], [2,1], [3, 3].

In the second example, the only possible array is [2, 2, 2].
import java.util.*;
import java.io.*;
public class A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            String s1[]=br.readLine().split(" ");
            int n=Integer.parseInt(s1[0]);
            int l=Integer.parseInt(s1[1]);
            int r=Integer.parseInt(s1[2]);
            
            long b[]=new long[3];
            b[0]=(long)Math.floor((double)r/3)-(long)Math.ceil((double)l/3)+1;
            b[1]=(long)Math.floor((double)(r-1)/3)-(long)Math.ceil((double)(l-1)/3)+1;
            b[2]=(long)Math.floor((double)(r-2)/3)-(long)Math.ceil((double)(l-2)/3)+1;
            
            long mod=1000000007;
            long a[][]=new long[n][3];
            a[0][0]=b[0];
            a[0][1]=b[1];
            a[0][2]=b[2];
            
            for(int i=1;i<n;i++)
            {
                  for(int j=0;j<3;j++)
                  {
                        for(int k=0;k<3;k++)
                        {
                              a[i][(j+k)%3]=(a[i][(j+k)%3]+(a[i-1][j]*b[k])%mod)%mod;
                        }
                  }
            }
            
            System.out.println(a[n-1][0]);
      }
}