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.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long l = sc.nextLong();
        long r = sc.nextLong();
        long m = 1000000007;
        long a = 0, b = 0, c = 0;

        while (l % 3 != 0) {
            if (l % 3 == 1) b++;
            if (l % 3 == 2) c++;
            l++;
        }

        while (r % 3 != 0) {
            if (r % 3 == 1) b++;
            if (r % 3 == 2) c++;
            r--;
        }

        a += (r - l) / 3 + 1;
        b += (r - l) / 3;
        c += (r - l) / 3;

        long a1 = a, b1 = b, c1 = c;
        long a2, b2, c2;
        for (int i = 2; i <= n; i++) {
            a2 = ((a1 * a) % m + (b1 * c) % m + (c1 * b) % m) % m;
            b2 = ((a1 * b) % m + (b1 * a) % m + (c1 * c) % m) % m;
            c2 = ((a1 * c) % m + (b1 * b) % m + (c1 * a) % m) % m;

            a1 = a2;
            b1 = b2;
            c1 = c2;
        }
        System.out.println(a1);
    }
}
