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.io.*;
import java.util.*;
public class Solution
{
    static int mod=(int)1e9+7;
    public static void main(String ag[])throws Exception
    {
        Scanner sc=new Scanner(System.in);
        int i,j,k;
        int N=sc.nextInt();
        int L=sc.nextInt();
        int R=sc.nextInt();
        int mod=(int)1e9+7;
        
        long m0=count(0,3,R)-count(0,3,L-1);
        long m1=count(1,3,R)-count(1,3,L-1);
        long m2=count(2,3,R)-count(2,3,L-1);
        
        //System.out.println(m0+" "+m1+" "+m2);
        long dp[][]=new long[N+1][3];
        dp[1][0]=m0;
        dp[1][1]=m1;
        dp[1][2]=m2;
        
        for(i=2;i<=N;i++)
        {
            dp[i][0]=add(mul(dp[i-1][0],m0),mul(dp[i-1][1],m2),mul(dp[i-1][2],m1));
            dp[i][1]=add(mul(dp[i-1][0],m1),mul(dp[i-1][1],m0),mul(dp[i-1][2],m2));
            dp[i][2]=add(mul(dp[i-1][0],m2),mul(dp[i-1][2],m0),mul(dp[i-1][1],m1));
        }
        System.out.println(dp[N][0]);
    }
    public static long add(long a,long b,long c)
    {
        return (a%mod+b%mod+c%mod)%mod;
    }
    public static long mul(long a,long b)
    {
        return (a%mod*b%mod)%mod;
    }
    public static long count(int mod,int num,int x)
    {
        if(mod>x)
        return 0;
        int remove=mod>0?1:0;
        return (x-mod)/num+remove;
    }
}