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.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        long mod = (long)(1e9+7);
        int n = sc.nextInt();
        int L = sc.nextInt(), R = sc.nextInt();
        long[] rec = new long[3];
        for(int i=0;i<3&&i+L<=R;i++){
            int cur = (L%3+i)%3;
            rec[cur] = Math.max(0,(R-L-i)/3)+1;
        }
        long[][] dp = new long[n+1][3];
        dp[0][0] = 1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<3;j++){ // find dp[i][j]
                for(int k=0;k<3;k++){
                    dp[i][j] += (dp[i-1][k]*rec[(j+3-k)%3])%mod;
                    dp[i][j] %= mod;
                }
            }
        }
        System.out.println(dp[n][0]);
    }
}
