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].
//package jan19;

import java.util.Scanner;

public class AyoubLostArray {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		long n=scn.nextLong();
		long l=scn.nextLong();
		long r=scn.nextLong();
		// We need to find how many numbers leave remainder 0,1,2 respectively.
		long[] rem=new long[3];
		rem[0] = (long) ((r/3) - ((l-1)/3));
		rem[1] = (long) (((r+2)/3) - ((l+1)/3));
		rem[2] = (long) (((r+1)/3) - ((l)/3));
		// It is calculated
		long mod=(long)1e9+7;
		/*DP where row represent size of array and columns represent remainders of 0,1,2*/
		long[][] dp=new long[(int) (n+1)][3];
		dp[0][0]=1;
		
		for(int i=1; i<=n; i++)
		{
			/*When length of array is i and remainder is 0*/
			dp[i][0]=((dp[i-1][0]*rem[0]%mod)+(dp[i-1][1]*rem[2]%mod)+(dp[i-1][2]*rem[1]%mod))%mod;
			/*When length of array is i and remainder is 1*/
			dp[i][1]=((dp[i-1][0]*rem[1]%mod)+(dp[i-1][1]*rem[0]%mod)+(dp[i-1][2]*rem[2]%mod))%mod;
			/*When length of array is i and remainder is 2*/
			dp[i][2]=((dp[i-1][0]*rem[2]%mod)+(dp[i-1][1]*rem[1]%mod)+(dp[i-1][2]*rem[0]%mod))%mod;
		}
		/*Now for n=2,l=1,r=4 answer is of length 2 where sum of elements divisible by 3*/
		/*And it is at dp[2][0] length 2 and remainder 0*/
		/*{3,3},{1,2},{2,1},{2,4},{4,2} */
		System.out.println(dp[(int) n][0]);

	}

}
