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 AyoubAndLostArray 
{
    static boolean debug = false ;
    static int MOD = (int)1e9+7 ;
    public static void main(String args[]) throws Exception 
    {
        BufferedReader bro = new BufferedReader(new InputStreamReader(System.in)) ;
        String[] S = bro.readLine().split(" ") ;
        long n = Long.parseLong(S[0]) ;
        long l = Long.parseLong(S[1]) ;
        long r = Long.parseLong(S[2]) ;
        long[][] dp = new long[3][(int)n+1] ;
        
        long beta = (long)Math.floor((double)r/3L) ;
        long alpha = (long)Math.ceil((double)l/3L) ;
        long zeros = beta-alpha+1 ;
        long lol = -1,uol = -1 ;
        if((alpha-1)*3+1>=l) lol = (alpha-1)*3+1 ;
        else lol = alpha*3+1 ;
        if(beta*3+1<=r) uol = beta*3+1 ;
        else uol = (beta-1)*3+1 ;
        if(debug) System.out.println("lol :"+lol+" uol :"+uol+"::"+ (beta*3+1));
        long ones = (uol-lol)/3+1 ;
        long twos = (r-l+1)-zeros-ones ;
        if(debug)
        {
            System.out.println("zeros :"+zeros+" ones :"+ones+" twos :"+twos);
        }
        dp[0][1] = zeros ;
        dp[1][1] = ones ;
        dp[2][1] = twos ;
        for(int i=2;i<=n;i++)
        {
            dp[0][i] = ((dp[0][i-1]*zeros)%MOD+(dp[1][i-1]*twos)%MOD+(dp[2][i-1]*ones)%MOD)%MOD ;
            dp[1][i] = ((dp[0][i-1]*ones)%MOD+(dp[1][i-1]*zeros)%MOD+(dp[2][i-1]*twos)%MOD)%MOD ;
            dp[2][i] = ((dp[0][i-1]*twos)%MOD+(dp[1][i-1]*ones)%MOD+(dp[2][i-1]*zeros)%MOD)%MOD ;
        }
        System.out.println(dp[0][(int)n]);
        
    }
}