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.*;
public class code{
    public static long mod = 1000000007;
    public static long[] val = new long[3];
    public static void modifier(long[] dp,long l,long r){
        long[] t = new long[3];
        t[0] = dp[0]*val[0]+dp[1]*val[2]+dp[2]*val[1];
        t[1] = dp[0]*val[1]+dp[1]*val[0]+dp[2]*val[2];
        t[2] = dp[0]*val[2]+dp[1]*val[1]+dp[2]*val[0];
        dp[0] = t[0]%mod;
        dp[1] = t[1]%mod;
        dp[2] = t[2]%mod;
    }
    public static void calcule(long l,long r){
        val[(int)(l%3)] = (r-l)/3+1;
        l++;
        if(l>r) return;
        val[(int)(l%3)] = (r-l)/3+1;
        l++;
        if(l>r) return;
        val[(int)(l%3)] = (r-l)/3+1;
    }
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long l = sc.nextInt();
        long r = sc.nextInt();
        calcule(l,r);
        long[] dp = new long[3];
        dp[0] = val[0];
        dp[1] = val[1];
        dp[2] = val[2];
        for(int i=1;i<n;i++){
            modifier(dp,l,r);
        }
        System.out.println(dp[0]);
    }
}