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) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int l = in.nextInt();
		int r = in.nextInt();
		long mod = 1000000007;
		long[][] f = new long[n+1][3];
		//num数组，用于存放l与r范围内的数mod 3 为0，1，2的计数
		int[] num = new int[3];
		num[0] = r/3 - (l-1)/3;
		num[1] = (r+2)/3 - (l+1)/3;
		num[2] = (r+1)/3 - l/3;
		f[1][0] = num[0];
		f[1][1] = num[1];
		f[1][2] = num[2];
		if(n == 1) {
			System.out.println(num[0]);
		}else {
			for(int i = 2;i <= n;i++) {
				f[i][0] = (f[i-1][0]*num[0] + f[i-1][1]*num[2] + f[i-1][2]*num[1]) % mod;
				f[i][1] = (f[i-1][0]*num[1] + f[i-1][1]*num[0] + f[i-1][2]*num[2]) % mod;
				f[i][2] = (f[i-1][0]*num[2] + f[i-1][1]*num[1] + f[i-1][2]*num[0]) % mod;
			}
			System.out.println(f[n][0]);
		}
	}

}