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].
// Created by Whiplash99
import java.io.*;
import java.util.*;
public class C
{
    private static final long MOD=1000000007;
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,j,N,l,r;
        long m[]=new long[3];

        String s[]=br.readLine().trim().split(" ");
        N=Integer.parseInt(s[0]);
        l=Integer.parseInt(s[1]);
        r=Integer.parseInt(s[2]);

        m[0]=(long)(Math.floor((double)(r)/3)-Math.ceil((double)(l)/3))+1;
        m[1]=(long)(Math.floor((double)(r-1)/3)-Math.ceil((double)(l-1)/3))+1;
        m[2]=(long)(Math.floor((double)(r-2)/3)-Math.ceil((double)(l-2)/3))+1;

        long dp[][]=new long[N+10][3];

        dp[0][0]=1;

        for(i=1;i<=N;i++)
        {
            for(j=0;j<3;j++)
            {
                dp[i][j]=(dp[i][j]+(m[0]*dp[i-1][j])%MOD)%MOD;
                dp[i][j]=(dp[i][j]+(m[1]*dp[i-1][(j+2)%3])%MOD)%MOD;
                dp[i][j]=(dp[i][j]+(m[2]*dp[i-1][(j+1)%3])%MOD)%MOD;
            }
        }

        System.out.println(dp[N][0]);
    }
}