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 java.lang.*;
public class ayoub{
   public static long mod = 1000000007;
   public static void main(String[] arg) throws IOException {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));    
      PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));      
      StringTokenizer st = new StringTokenizer(br.readLine());
      int n = Integer.parseInt(st.nextToken());
      int l = Integer.parseInt(st.nextToken());
      int r = Integer.parseInt(st.nextToken());
      int countzero = 0;
      int countone = 0;
      int counttwo = 0;
      int range = r-l+1;
      if(range%3 == 0){
         countzero = range/3;
         counttwo = range/3;
         countone = range/3;
      }
      else if(range%3 == 1){
         if(l%3 == 0){
            countzero++;
         }
         else if(l%3 == 1){
            countone++;
         }
         else{
            counttwo++;
         }
         countzero += range/3;
         counttwo += range/3;
         countone += range/3;
      }
      else{
         if(l%3 == 0){
            countzero++;
            countone++;
         }
         else if(l%3 == 1){
            countone++;
            counttwo++;
         }
         else{
            counttwo++;
            countzero++;
         }
         countzero += range/3;
         counttwo += range/3;
         countone += range/3;
      }
      long[][] dp = new long [n][3];
      dp[0][0] = countzero%mod;
      dp[0][1] = countone%mod;
      dp[0][2] = counttwo%mod;
      for(int k = 1; k < n; k++){
         dp[k][0] = (((countzero*dp[k-1][0])%mod) + ((counttwo*dp[k-1][1])%mod) + ((countone*dp[k-1][2])%mod))%mod;
         dp[k][1] = (((countone*dp[k-1][0])%mod) + ((countzero*dp[k-1][1])%mod) + ((counttwo*dp[k-1][2])%mod))%mod;
         dp[k][2] = (((counttwo*dp[k-1][0])%mod) + ((countone*dp[k-1][1])%mod) + ((countzero*dp[k-1][2])%mod))%mod;
      }
      out.println(dp[n-1][0]);
      out.close();
   }
}