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.io.*;

public class Main {

    final static long mo = 1000000007;

    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

        //读入
        long n, l, r;
        in.nextToken();
        n = (long) in.nval;
        in.nextToken();
        l = (long) in.nval;
        in.nextToken();
        r = (long) in.nval;

        long dp[][] = new long[(int) n + 1][3];

        long mo0, mo1, mo2;
        mo0 = (r / 3 - (l - 1) / 3) % mo;
        mo1 = ((r + 2) / 3 - (l + 1) / 3) % mo;
        mo2 = ((r + 1) / 3 - l / 3) % mo;

        dp[1][0] = mo0;//区间中余3为0的个数
        dp[1][1] = mo1;//区间中余3为1的个数
        dp[1][2] = mo2;//区间中余3为2的个数


        //到dp[i][j]的时候前i个数的和对3取余结果为j的方案数
        for (int i = 2; i <= n; i++) {
            dp[i][0] = ((dp[i - 1][0] * mo0) % mo + (dp[i - 1][1] * mo2) % mo + (dp[i - 1][2] * mo1) % mo) % mo;
            dp[i][1] = ((dp[i - 1][0] * mo1) % mo + (dp[i - 1][1] * mo0) % mo + (dp[i - 1][2] * mo2) % mo) % mo;
            dp[i][2] = ((dp[i - 1][0] * mo2) % mo + (dp[i - 1][1] * mo1) % mo + (dp[i - 1][2] * mo0) % mo) % mo;
        }

        out.println(dp[(int) n][0]);


        out.flush();
    }
}
 		      	 	  	  		  		 	 	   	