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 Solution {


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int l = sc.nextInt();
        int r = sc.nextInt();
        long MOD = 1000000007;
        int[] val = new int[3];
        if (l%3 == 1){
            val[1]++;
            l++;
        }
        if (l <= r && l%3 == 2){
            val[2]++;
            l++;
        }



        if (l <= r){
            int len = r - l + 1;
            val[0] += (int)Math.ceil(len/3.0);
            val[1] += Math.max(0, (int)Math.ceil((len-1)/3.0));
            val[2] += Math.max(0, (int)Math.ceil((len-2)/3.0));
        }

        int[] dp = new int[3];

        for (int i=0; i<n; i++){

            long[] temp = new long[3];
            if (i == 0){
                temp[0] = (long) val[0];
                temp[1] = (long) val[1];
                temp[2] = (long) val[2];
            }else {
                temp[0] = (long)dp[0]*(long)val[0] + (long)dp[1]*(long)val[2] + (long)dp[2]*(long)val[1];
                temp[1] = (long)dp[0]*(long)val[1] + (long)dp[1]*(long)val[0] + (long)dp[2]*(long)val[2];
                temp[2] = (long)dp[0]*(long)val[2] + (long)dp[2]*(long)val[0] + (long)dp[1]*(long)val[1];
            }

            for (int j=0; j<3; j++) {
                dp[j] = (int) (temp[j] % MOD);
            }
        }

        System.out.println(dp[0]);

    }
}


