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.math.BigInteger;
import java.util.Scanner;

/**
 * Created by andre on 1/29/2019.
 */
public class SecondTry {

    static long modulus;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int l = in.nextInt();
        int r = in.nextInt();


        modulus = 1000000007;

        // First determine how many integers there are between l and r (inclusive)
        // that are of each equivalence class mod 3
        int num_zero, num_one, num_two;
        num_zero = num_one = num_two = (r + 1 - l)/3;
        if((l + 2) % 3 == r % 3){
            // Do Nothing, numbers are already correct
        }
        else if((l + 1) % 3 == r % 3) {
            switch (l % 3){
                case 0:
                    num_zero++;
                    num_one++;
                    break;
                case 1:
                    num_one++;
                    num_two++;
                    break;
                case 2:
                    num_zero++;
                    num_two++;
            }
        }
        else{
            switch (l % 3){
                case 0:
                    num_zero++;
                    break;
                case 1:
                    num_one++;
                    break;
                case 2:
                    num_two++;
            }
        }

        long[][] res = new long[n][3];
        res[0][0] = num_zero;
        res[0][1] = num_one;
        res[0][2] = num_two;
        long tmp;

        for(int i = 1; i < n; i++){
            res[i][0] = res[i-1][0] * num_zero + res[i-1][1] * num_two + res[i-1][2] * num_one;
            res[i][1] = res[i-1][0] * num_one + res[i-1][1] * num_zero + res[i-1][2] * num_two;
            res[i][2] = res[i-1][0] * num_two + res[i-1][1] * num_one + res[i-1][2] * num_zero;
            res[i][0] %= modulus;
            res[i][1] %= modulus;
            res[i][2] %= modulus;
        }

        System.out.println(res[n-1][0]);
    }
}
