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.Arrays;
import java.util.Scanner;

public class Solution {
	static int mod = 1000000007;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt(), l = scan.nextInt(), r = scan.nextInt();
		long[][] arr = new long[n][3];
		arr[0][0] = getLR(l,r,0);
		arr[0][1] = getLR(l,r,1);
		arr[0][2] = getLR(l,r,2);
		for (int i=1;i<n;i++) {
			arr[i][0]=((arr[0][0] * arr[i-1][0])%mod +  (arr[i-1][1] * arr[0][2])%mod + (arr[0][1] * arr[i-1][2])%mod) % mod;
			arr[i][1]=((arr[0][2]*arr[i-1][2])%mod +  (arr[i-1][0] * arr[0][1])%mod +  (arr[0][0] * arr[i-1][1])%mod) % mod;
			arr[i][2]=((arr[0][1]*arr[i-1][1])%mod +  (arr[i-1][0] * arr[0][2]) %mod+  (arr[0][0] * arr[i-1][2])%mod) % mod;
		}
		System.out.println(arr[n-1][0]);
		scan.close();
	}

	private static long getC(long i) {
		return (i*(i-1) + i) % mod;
	}

	private static long getLR(int l, int r, int mod) {
		while(l%3!=mod&&l<=r)
			l++;
		while(r%3!=mod&&r>=l)
			r--;
		if(r>=l)return ((r-l)/3)+1;
		else return 0;
	}
}
