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.*;
import java.util.*;

public class Codeforces1105C {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int l = Integer.parseInt(st.nextToken());
		int r = Integer.parseInt(st.nextToken());
		
		long[] mod = new long[3];
		mod[0] = (long) ((r/3) - ((l-1)/3));
		mod[1] = (long) (((r+2)/3) - ((l+1)/3));
		mod[2] = (long) (((r+1)/3) - ((l)/3));
				
		int[][] ans = new int[n+1][3];
		ans[0][0] = 1;
		
		for (int i = 1; i <= n; i++) {
			long val = (long) ans[i-1][0] * mod[0] + (long) ans[i-1][1] * mod[2] + (long) ans[i-1][2] * mod[1];
			ans[i][0] = (int) (val%1000000007);
			val = (long) ans[i-1][0] * mod[1] + (long) ans[i-1][1] * mod[0] + (long) ans[i-1][2] * mod[2];
			ans[i][1] = (int) (val%1000000007);
			val = (long) ans[i-1][0] * mod[2] + (long) ans[i-1][1] * mod[1] + (long) ans[i-1][2] * mod[0];
			ans[i][2] = (int) (val%1000000007);
		}
		
		System.out.println(ans[n][0]);
	}
}