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.*;

public class Main {
	public static void main(String[] args)
	{
		//p_1098C();
		//p_445A();
		//p_431C();
		//p_706C();
		//p_44E();
		//p_106C();
		//p_1105A();
		//p_1105B();
		p_1105C();

	}
	private static void p_1105C()
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int l = sc.nextInt();
		int r = sc.nextInt();
		long[][] sumA = new long[n][3];
		long number = 1000000007;
		for( int i=0;i<n;i++)
		{
			long num0 = r/3 - (l-1)/3;
			long num1 = r/3 + (r%3>0?1:0) - ((l-1)/3 + ((l-1)%3>0?1:0));
			long num2 = r/3 + (r%3==2?1:0) - ((l-1)/3 + ((l-1)%3==2?1:0));
			if(i==0)
			{
				sumA[i][0] = num0;
				sumA[i][1] = num1;
				sumA[i][2] = num2;
			}
			else
			{
				sumA[i][0] = (sumA[i-1][0] * num0 + sumA[i-1][2] * num1 + sumA[i-1][1] * num2)%number;
				sumA[i][1] = (sumA[i-1][0] * num1 + sumA[i-1][2] * num2 + sumA[i-1][1] * num0)%number;
				sumA[i][2] = (sumA[i-1][1] * num1 + sumA[i-1][0] * num2 + sumA[i-1][2] * num0)%number;
			}
		}
		System.out.println(sumA[n-1][0]);
	}
}