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.math.BigInteger;
import java.text.DecimalFormat;
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer stk = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(stk.nextToken());
        int l = Integer.parseInt(stk.nextToken());
        int r = Integer.parseInt(stk.nextToken());

        int b[] = new int[3];
        b[0] = r/3 - (l-1)/3;
        b[1] = (r+1)/3 - (l)/3;
        b[2] = (r+2)/3 - (l+1)/3;
        long[][] dp = new long[n+1][3];
        for(int i=0;i<3;i++)dp[1][i] = b[i];
        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]*b[k]%1000000007;
                    dp[i][(j+k)%3] %= 1000000007;
                }
            }
        }
        System.out.println(dp[n][0]);
    }
}
