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 {
    private static int[][] dp;
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        dp = new int [200005][];
        int n,l,r;
        final int mod = 1000000007;
        n = cin.nextInt();
        l = cin.nextInt();
        r = cin.nextInt();
        dp[0] = new int [3];
        dp[0][0] = 1;
      //  dp[0][1] = 1;
    //    dp[0][2] = 1;
        long c = r - l;
        c = c - c%3;
        c = r - c;
        int []cnt = new int[3];
        for(int i = l ; i <= c ; ++i){
          //  System.out.println(i);
            ++cnt[(i%3)];
        }
        long dep = (r-l)/3;
        for(int i = 1 ; i <= n ; ++i){
            dp[i] = new int[3];
            for(int j = 0 ; j < 3 ; ++j){
                for(int k = 0 ; k < 3 ; ++k){
                    dp[i][(j+k)%3] = (int)((dp[i][(j+k)%3] + ((dep+cnt[j])*dp[i-1][k]%mod))%mod);
                }
            }
        }
        System.out.println(dp[n][0]);
    }
}
