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 C {

   public static final long mod = 1000000007;
   
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      
      int n = sc.nextInt();
      long l = sc.nextLong();
      long r = sc.nextLong();
      
      long[][] dp = new long[n+1][3];
      for(int i = 0; i < 3; i++) {
         dp[1][i] = (long)Math.floor((r-i)/3.0) - (long)Math.ceil((l-i)/3.0) + 1;
      }
      for(int i = 2; i <= n; i++) {
         for(int j = 0; j < 3; j++) {
            for(int k = 0; k < 3; k++) {
               dp[i][(j+k)%3] += dp[i-1][j] * dp[1][k];
               dp[i][(j+k)%3] %= mod;
            } 
         }
      }
      System.out.println(dp[n][0]);
   }
}