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 codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Graph
{  
	public static void main (String[] args) throws java.lang.Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s[] = br.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        int l = Integer.parseInt(s[1]);
        int r = Integer.parseInt(s[2]);
        
        int zeros = 0;
        int ones = 0;
        int twos = 0;
        int mod = 1000000007;
        
        while(l <= r && l % 3 != 0) {
            if(l % 3 == 1)
                ones += 1;
            else 
                twos += 1;
            
            l += 1;
        }
        
        while(r >= l && r % 3 != 2) {
            if(r % 3 == 0)
                zeros += 1;
            else 
                ones += 1;
                
            r -= 1;
        }
        
        int len = r - l + 1;
        
        if(len > 0) {
            zeros += len / 3;
            ones += len / 3;
            twos += len / 3;
        }
        
        //System.out.println(zeros + " " + ones + " " + twos);
        
        long dp[][] = new long[3][n];
        
        dp[0][0] = zeros;
        dp[1][0] = ones;
        dp[2][0] = twos;
        
        for(int i = 1; i < n; i++) {
            dp[0][i] = ((dp[0][i - 1] * zeros) % mod + (dp[1][i - 1] * twos) % mod + (dp[2][i - 1] * ones) % mod) % mod;
            dp[1][i] = ((dp[0][i - 1] * ones) % mod + (dp[1][i - 1] * zeros) % mod + (dp[2][i - 1] * twos) % mod) % mod;
            dp[2][i] = ((dp[0][i - 1] * twos) % mod + (dp[1][i - 1] * ones) % mod + (dp[2][i - 1] * zeros) % mod) % mod;
        
            //System.out.println(dp[0][i]);
        }
        
        System.out.println(dp[0][n - 1]);
	}    
}
