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.util.*;
public class p1109C {
    public static void main(String[] args) {
        Scanner scn=new Scanner(System.in);
        long n=scn.nextInt();
        long l=scn.nextInt();
        long r=scn.nextInt();
        long m=1000000000+7;
        long[] arr=new long[3];
        for(int i=0;i<3;i++){
            long v1=1+(l-1-i)/3;
            long v2=1+(r-i)/3;
            if(r<i){
                v2=0;
            }
            if((l-1)<i){
                v1=0;
            }
            arr[i]=v2-v1;
        }
        
        long[] ans=new long[3];
        ans[0]=arr[0];
        ans[1]=arr[1];
        ans[2]=arr[2];
        for(int i=1;i<n;i++){
            long v1=ans[0]*arr[0]+ans[1]*arr[2]+ans[2]*arr[1];
            long v2=ans[0]*arr[1]+ans[1]*arr[0]+ans[2]*arr[2];
            long v3=ans[0]*arr[2]+ans[1]*arr[1]+ans[2]*arr[0];
            ans[0]=v1%m;
            ans[1]=v2%m;
            ans[2]=v3%m;
        }
        System.out.println(ans[0]);
    }
}
