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 C {
    public static void main(String[] aaaa) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int l = scanner.nextInt();
        int r = scanner.nextInt();
        long[][] dp = new long[200005][3];
        long mod = 1000000007;
        dp[0][0] = 1;
        int a = (r / 3) - ((l - 1) / 3);
        int b = ((r + 2) / 3) - ((l + 1) / 3);
        int c = ((r + 1) / 3) - ((l) / 3);

//        System.out.println(a);
//        System.out.println(b);
//        System.out.println(c);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 3; j++) {
                dp[i + 1][j] += dp[i][j] * a;
                dp[i + 1][j] %= mod;
                dp[i + 1][(j + 1) % 3] += dp[i][j] * b;
                dp[i + 1][(j + 1) % 3] %= mod;
                dp[i + 1][(j + 2) % 3] += dp[i][j] * c;
                dp[i + 1][(j + 2) % 3] %= mod;
            }
        }
        System.out.println((dp[n][0]) % mod);




    }
}
