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.Scanner;

public class ProblemC {
    static final long MOD = 1000000007;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int l = sc.nextInt();
		int r = sc.nextInt();
		sc.close();
		long[][] dp = new long[n + 1][3];
		int []k = new int[3];
		for(int i = 0; i < 3; i++) {
			k[i] = (r - i + 3) / 3 - (l - i + 2) / 3;
		}
		dp[1][0] = k[0];
		dp[1][1] = k[1];
		dp[1][2] = k[2];
		for(int i = 2; i <= n; i++) {
			dp[i][0] = dp[i-1][0] * k[0] + dp[i-1][1] * k[2] + dp[i-1][2] * k[1];
			dp[i][1] = dp[i-1][0] * k[1] + dp[i-1][1] * k[0] + dp[i-1][2] * k[2];
			dp[i][2] = dp[i-1][0] * k[2] + dp[i-1][1] * k[1] + dp[i-1][2] * k[0];
			dp[i][0] %= MOD;
			dp[i][1] %= MOD;
			dp[i][2] %= MOD;
		}
		System.out.println(dp[n][0]);
	}
}