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 Main {

	public static void main(String[] args) {
		long[][] tab = new long[200000][3];
		long max = 1000000007;
		long n,l,r;
		Scanner in = new Scanner(System.in);
		n = in.nextLong();
		l = in.nextLong();
		r = in.nextLong();
		
		tab[0][0] = (r-l+1)/3;
		tab[0][1] = tab[0][0];
		tab[0][2] = tab[0][0];
		long v = l%3;
		for(long i = 0 ; i < (r-l+1)%3 ; i++,v++) {
			tab[0][(int) v%3]++;
		}
		
		for(int i = 1 ; i < n ; i++) {
			tab[i][0] = ((tab[i-1][0] * tab[0][0]%max + tab[i-1][1] * tab[0][2]%max)%max + tab[i-1][2] * tab[0][1]%max)%max;
			tab[i][1] = ((tab[i-1][0] * tab[0][1]%max + tab[i-1][1] * tab[0][0]%max)%max + tab[i-1][2] * tab[0][2]%max)%max;
			tab[i][2] = ((tab[i-1][0] * tab[0][2]%max + tab[i-1][1] * tab[0][1]%max)%max + tab[i-1][2] * tab[0][0]%max)%max;
		}
		
		System.out.println(tab[(int) (n-1)][0]);
	}
}
