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 long dp[][];
    static int N;
    static long m0,m1,m2;
    static int L,R;
    static int mod=(int)1e9+7;
    public static void main(String ag[])
    {
        Scanner sc=new Scanner(System.in);
        N=sc.nextInt();
        L=sc.nextInt();
        R=sc.nextInt();
        
        dp=new long[N][3];
        for(int i=0;i<N;i++)
        Arrays.fill(dp[i],-1);
        
        m0=count(0,3,R)-count(0,3,L-1);
        m1=count(1,3,R)-count(1,3,L-1);
        m2=count(2,3,R)-count(2,3,L-1);
        
        long ans=0;
        ans=(ans%mod+(find(1,0)%mod*m0%mod))%mod;
        ans%=mod;
        ans=(ans%mod+(find(1,1)%mod*m1%mod))%mod;
        ans%=mod;
        ans=(ans%mod+(find(1,2)%mod*m2%mod))%mod;
        System.out.println(ans);
    }
    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;
    }
    public static long find(int id,int sum)
    {
        if(id==N)
        {
            if(sum==0)
            return 1;
            return 0;
        }
        
        if(dp[id][sum]!=-1)
        return dp[id][sum];
        
        long ans=0;
        ans=(ans%mod+(m0%mod*find(id+1,(0+sum)%3)))%mod;
        ans%=mod;
        ans=(ans%mod+(m1%mod*find(id+1,(1+sum)%3)))%mod;
        ans%=mod;
        ans=(ans%mod+(m2%mod*find(id+1,(2+sum)%3)))%mod;
        
        return dp[id][sum]=ans%mod;
        
    }
}