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.lang.*;
import java.io.*;
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
	    Scanner sc=new Scanner(System.in);
	   // BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    int t=1;
	   // t=sc.nextInt();
	    //int t=Integer.parseInt(br.readLine());
	    while(--t>=0){
	        int n=sc.nextInt();
	        int l=sc.nextInt();
	        int r=sc.nextInt();
	        
	        long dp[][]=new long[n][3];
	        long M=1000000007;
	        long r0=(long)(Math.floor((double)(r)/3.0)-Math.ceil((double)(l)/3.0))+1;
	        long r1=(long)(Math.floor((double)(r-1.0)/3.0)-Math.ceil((double)(l-1.0)/3.0))+1;
	        long r2=(long)(Math.floor((double)(r-2.0)/3.0)-Math.ceil((double)(l-2.0)/3.0))+1;
	        dp[0][0]=r0;
	        dp[0][1]=r1;
	        dp[0][2]=r2;
	        for(int i=1;i<n;i++){
	            dp[i][0]=((dp[i-1][0]*r0)%M+(dp[i-1][2]*r1)%M+(dp[i-1][1]*r2)%M)%M;
	            dp[i][1]=((dp[i-1][1]*r0)%M+(dp[i-1][0]*r1)%M+(dp[i-1][2]*r2)%M)%M;
	            dp[i][2]=((dp[i-1][0]*r2)%M+(dp[i-1][2]*r0)%M+(dp[i-1][1]*r1)%M)%M;
	            
	             }
	        System.out.println(dp[n-1][0]);
	        
	     
	        
	    }
	    
	}
}