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].
/*
roses are red
memes are neat
all my test cases time out
lmao yeet
*/
import java.util.*;
import java.io.*;

   public class C
   {
      static long MOD = 1000000007L;
      public static void main(String args[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int L = Integer.parseInt(st.nextToken());
         int R = Integer.parseInt(st.nextToken());
         //
         long remain = ((long)N*L)%3;
         if(L == R)
         {
            if(remain != 0)
               System.out.println(0);
            else
               System.out.println(1);
            return;
         }
         long[][] dp = new long[N+1][3];
         int[] count = new int[3];
         for(int i=0; i < 3; i++)
         {
            count[i] = calc(L, R, i);
            dp[0][i] = count[i];
         }
         for(int i=0; i < N; i++)
         {
            dp[i+1][0] = dp[i][0]*count[0];
            dp[i+1][0] %= MOD;
            dp[i+1][0] += dp[i][1]*count[2];
            dp[i+1][0] %= MOD;
            dp[i+1][0] += dp[i][2]*count[1];
            dp[i+1][0] %= MOD;
            //
            dp[i+1][1] = dp[i][1]*count[0];
            dp[i+1][1] %= MOD;
            dp[i+1][1] += dp[i][0]*count[1];
            dp[i+1][1] %= MOD;
            dp[i+1][1] += dp[i][2]*count[2];
            dp[i+1][1] %= MOD;
            //
            dp[i+1][2] = dp[i][0]*count[2];
            dp[i+1][2] %= MOD;
            dp[i+1][2] += dp[i][2]*count[0];
            dp[i+1][2] %= MOD;
            dp[i+1][2] += dp[i][1]*count[1];
            dp[i+1][2] %= MOD;
         }
         System.out.println(dp[N-1][0]%MOD);
      }
      public static int calc(int A, int B, int M)
      {
         A += M;
         B += M;
         if (A % 3 == 0) 
            return (B / 3) - (A / 3) + 1; 
        return (B / 3) - (A / 3); 
      }
   }