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.*;
import java.lang.*;
public class test{
	public static void main(String[] args){
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int l=s.nextInt();
		int r=s.nextInt();
		int x0=0,y1=0,z2=0;
		 while(l%3!=0)
        {
            if(l%3==1)
                y1++;
            if(l%3==2)
                z2++;
            l++;
        }
        while(r%3!=0)
        {
            if(r%3==1)
                y1++;
            if(r%3==2)
                z2++;
            r--;
        }
        x0=(r-l)/3+1;
        y1+=(r-l)/3;
        z2+=(r-l)/3;
		long m=(long)1e9+7;
		long[][] dp=new long[n+1][3];
		dp[1][0]=x0;
		dp[1][1]=y1;
		dp[1][2]=z2;
		for(int i=2;i<=n;i++){
			dp[i][0]=((dp[i-1][0]*x0)%m+(dp[i-1][1]*z2)%m+(dp[i-1][2]*y1)%m)%m;
			dp[i][1]=((dp[i-1][0]*y1)%m+(dp[i-1][1]*x0)%m+(dp[i-1][2]*z2)%m)%m;
			dp[i][2]=((dp[i-1][0]*z2)%m+(dp[i-1][1]*y1)%m+(dp[i-1][2]*x0)%m)%m;
		}
		System.out.println(dp[n][0]);
	}
}