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 C{
	static Scanner fr = new Scanner(System.in);
	static int mod = (int)(1e9 + 7);
	public static void main(String args[]){
		int n = fr.nextInt();
		int l = fr.nextInt();
		int r = fr.nextInt();
		
		int rem[] = new int[3];
		rem[0] = r/3 - (l-1)/3;
		rem[1] = (int)Math.ceil((r)/3.0) - (int)Math.ceil((l-1)/3.0);
		rem[2] = (int)Math.ceil((r-1)/3.0) - (int)Math.ceil((l-2)/3.0);
		
		long dp[][] = new long[3][n];
		dp[0][0] = rem[2];
		dp[1][0] = rem[1];
		dp[2][0] = rem[0];
		
		for(int i=1;i<n;i++){
			dp[0][i] = ((dp[1][i-1] * dp[1][0]) % mod + (dp[2][i-1] * dp[0][0]) % mod + 
						(dp[0][i-1] * dp[2][0]) % mod) % mod;
			
			dp[1][i] = ((dp[1][i-1] * dp[2][0]) % mod + (dp[2][i-1] * dp[1][0]) % mod + 
					(dp[0][i-1] * dp[0][0]) % mod) % mod;
			
			dp[2][i] = ((dp[1][i-1] * dp[0][0]) % mod + (dp[2][i-1] * dp[2][0]) % mod + 
					(dp[0][i-1] * dp[1][0]) % mod) % mod;
		}
		
		System.out.println(dp[2][n-1]);
		
	}
	
	static int[] nextArray(int n){
		int arr[] = new int[n];
		for(int i=0;i<n;i++){
			arr[i] = fr.nextInt();
		}
		return arr;
	}
}
