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 C1105 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int l = scanner.nextInt();
        int r = scanner.nextInt();
        //区间内模3余0\1\2的数的个数
        int mod0 = r/3-(l-1)/3;
        int mod1 = (r+1)/3-l/3;
        int mod2 = (r+2)/3-(l+1)/3;
        long[][] arr = new long[n+1][3];
        for(int i=1;i<=n;i++){
            if(i == 1){
                arr[i][0] = mod0;
                arr[i][1] = mod1;
                arr[i][2] = mod2;
            }
            else{
                arr[i][0] = arr[i-1][0]*mod0 + arr[i-1][1]*mod2 + arr[i-1][2]*mod1;
                arr[i][0] = arr[i][0]%(1000000007);
                arr[i][1] = arr[i-1][0]*mod1 + arr[i-1][1]*mod0 + arr[i-1][2]*mod2;
                arr[i][1] = arr[i][1]%(1000000007);
                arr[i][2] = arr[i-1][0]*mod2 + arr[i-1][1]*mod1 + arr[i-1][2]*mod0;
                arr[i][2] = arr[i][2]%(1000000007);
            }
        }

        System.out.println(arr[n][0]);

    }
}
