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.Scanner;

public class C1105 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int l=s.nextInt();
		int r=s.nextInt();
		long [][] dp=new long [n+1][4];
		long k=1000000007;
		dp[1][1]=0;
		dp[1][2]=0;
		dp[1][3]=0;
		if(l%3==0)
		{
			dp[1][3]=((r-l)/3)+1;
			if(r-l>=2)
			dp[1][2]=((r-(l+2))/3)+1;
			if(r-l>=1)
			dp[1][1]=((r-(l+1))/3)+1;
		}
		else if(l%3==1)
		{
			dp[1][1]=((r-l)/3)+1;
			if(r-l>=1)
			dp[1][2]=((r-(l+1))/3)+1;
			if(r-l>=2)
			dp[1][3]=((r-(l+2))/3)+1;
		}
		else
		{
			dp[1][2]=(((r-l)/3)+1);
			if(r-l>=1)
			dp[1][3]=((r-(l+1))/3)+1;
			if(r-l>=2)
			dp[1][1]=((r-(l+2))/3)+1;
		}
		long a=Math.max(0, dp[1][1])%k;
		long b=Math.max(0, dp[1][2])%k;
		long c=Math.max(0, dp[1][3])%k;
		dp[1][1]=a;
		dp[1][2]=b;
		dp[1][3]=c;
	//	System.out.println(a+" "+b+" "+c);
		for(int i=2;i<n+1;i++)
		{
			dp[i][1]=(((dp[i-1][1]*c)%k+(dp[i-1][2]*b)%k)%k+(dp[i-1][3]*a)%k)%k;
			dp[i][2]=(((dp[i-1][1]*a)%k+(dp[i-1][2]*c)%k)%k+(dp[i-1][3]*b)%k)%k;
			dp[i][3]=(((dp[i-1][1]*b)%k+(dp[i-1][2]*a)%k)%k+(dp[i-1][3]*c)%k)%k;
		}
		System.out.println(dp[n][3]);
		
	}

}