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 Codechef
{
	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 m=1000000007l;
	        long dp[][]=new long[n][3];
	        for(int i=0;i<n;i++) {
	        	long t0=((r-l)/3l+1l);
	        	long t1=(r-l-1l)/3l+1l;
	        	long t2=(r-l-2l)/3l+1l;
	        	if((r-l-1)<0)
	        		t1=0l;
	        	if((r-l-2)<0)
	        		t2=0l;
	        	long l0,l1,l2;
	        	if(l%3==0) {
	        		l0=t0;l1=t1;l2=t2;
	        	}
	        	else if(l%3==1) {
	        		l0=t2;l1=t0;l2=t1;
	        	}
	        	else {
	        		l0=t1;l1=t2;l2=t0;
	        	}
	        	if(i==0) {
	        		dp[i][0]=l0;dp[i][1]=l1;dp[i][2]=l2;
	        		//System.out.println(dp[i][0]+" "+dp[i][1]+" "+dp[i][2]);
	        		continue;
	        	}
	        		dp[i][0]=((dp[i-1][0]*l0)%m+(dp[i-1][1]*l2)%m+(dp[i-1][2]*l1)%m)%m;
	        		dp[i][1]=((dp[i-1][0]*l1)%m+(dp[i-1][1]*l0)%m+(dp[i-1][2]*l2)%m)%m;
	        		dp[i][2]=((dp[i-1][0]*l2)%m+(dp[i-1][1]*l1)%m+(dp[i-1][2]*l0)%m)%m;
	        	 //   System.out.println(dp[i][0]+" "+dp[i][1]+" "+dp[i][2]);
	        	
	        }
	        System.out.println(dp[n-1][0]);
	    }
	    
	}
}