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].
//package test;

import java.util.Scanner;

public class starter
{
	
	static int MOD = 1000000007 ;
	
		public static void main ( String args[] )
			throws Exception 
		{
			Scanner conio = new Scanner ( System.in ) ;
		
			int n = conio.nextInt() ;
			int l = conio.nextInt() ;
			int r = conio.nextInt() ;
			
			
			int dp[][] = new int [n+1][3] ;
			
			// calculate base cases 
			
			int m0 = dp[1][0] = Math.floorDiv( r , 3 ) - ( int ) Math.ceil( l / 3.0 ) + 1  ;
			int m1 = dp[1][1] = Math.floorDiv( r - 1 , 3 ) - ( int ) Math.ceil( (l - 1) / 3.0 ) + 1  ;
			int m2 = dp[1][2] = Math.floorDiv( r - 2 , 3 ) - ( int ) Math.ceil( (l - 2) / 3.0 ) + 1 ;
			
			
			for ( int i = 2 ; i < n+1 ; ++i )
			{
				dp[i][0] = add(dp[i][0], mult(dp[i - 1][0], m0));
			    dp[i][0] = add(dp[i][0], mult(dp[i - 1][1], m2));
			    dp[i][0] = add(dp[i][0], mult(dp[i - 1][2], m1));
			    dp[i][1] = add(dp[i][1], mult(dp[i - 1][0], m1));
			    dp[i][1] = add(dp[i][1], mult(dp[i - 1][1], m0));
			    dp[i][1] = add(dp[i][1], mult(dp[i - 1][2], m2));
			    dp[i][2] = add(dp[i][2], mult(dp[i - 1][0], m2));
			    dp[i][2] = add(dp[i][2], mult(dp[i - 1][1], m1));
			    dp[i][2] = add(dp[i][2], mult(dp[i - 1][2], m0)); 
			}
			
			System.out.println( dp[n][0] ) ; 
			
			
			
			
			
			
			
			
			
			
			
			
			
			
			conio.close() ;  
			
		}
		
		private static int add ( long x1 , long x2 )
		{
			return ( int ) (((x1%MOD) + (x2%MOD)) % MOD ) ;
		}
		
		private static int mult ( long x1 , long x2 )
		{
			return ( int ) (((x1%MOD) * (x2%MOD)) % MOD ) ;
		}
		
		
}