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 P1105C {

	static int n, l, r;
	static long zeroR, oneR, twoR;
	static int MOD = (int)1e9+7;
	static long[][] dp = new long[200001][3];
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		l = in.nextInt();
		r = in.nextInt();
		for(int i=0; i<200001; i++) {
			for(int j=0; j<3; j++) {
				dp[i][j]=-1;
			}
		}
		zeroR = findNumbersDivisibleByThree(l, r);
		oneR = findNumbersDivisibleByThree(l-1, r-1);
		twoR = findNumbersDivisibleByThree(l-2, r-2);
		
//		System.out.println(zeroR+" "+oneR+" "+twoR);
		
		System.out.println(solve(0, 0));
	}
	
	public static long solve(int i, int remainder) {
		if (i==n) {
			if (remainder==0) return 1L;
			else return 0;
		}
		
		if (dp[i][remainder]!=-1) return dp[i][remainder];
		
		return dp[i][remainder]=((zeroR*1L*solve(i+1, remainder))%MOD+(oneR*1L*solve(i+1, (remainder+1)%3))%MOD+(twoR*1L*solve(i+1, (remainder+2)%3))%MOD)%MOD;
	}
	
	public static long findNumbersDivisibleByThree(int l, int r) {
		int last = 3*(int)Math.floor((double)r/3);
		int first = 3*(int)Math.ceil((double)l/3);
		return 1+(last-first)/3;
	}

}
