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.io.PrintWriter;
import java.util.Scanner;

public class Code_1105C_AyoubAndLostArray {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int n=in.nextInt();
        long l=in.nextLong(),r=in.nextLong();
        long MOD= (long) (1e9+7);
        int[] count=new int[3];
        if(l<=r) count[(int) (l % 3)] = (int) (r - l) / 3 + 1;
        if(l+1<=r) count[(int)((l+1)%3)] = (int)(r-(l+1))/3+1;
        if(l+2<=r) count[(int)((l+2)%3)] = (int)(r-(l+2))/3+1;
        long[][] dp=new long[n][3];
        dp[0][0]=count[0];
        dp[0][1]=count[1];
        dp[0][2]=count[2];
        for(int i=1;i<n;++i){
            dp[i][0]=dp[i-1][0]%MOD*count[0]%MOD+dp[i-1][1]%MOD*count[2]%MOD+dp[i-1][2]%MOD*count[1]%MOD;
            dp[i][1]=dp[i-1][0]%MOD*count[1]%MOD+dp[i-1][1]%MOD*count[0]%MOD+dp[i-1][2]%MOD*count[2]%MOD;
            dp[i][2]=dp[i-1][0]%MOD*count[2]%MOD+dp[i-1][1]%MOD*count[1]%MOD+dp[i-1][2]%MOD*count[0]%MOD;
        }
        System.out.println(dp[n-1][0]%MOD);
        out.close();
    }
}
