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 Main {
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		long[][] dp = new long[200007][3];
		int n = cin.nextInt();
		long l = cin.nextLong();
		long r = cin.nextLong();
		long three = r / 3 - (l - 1) / 3;
		long two = (r + 1) / 3 - l / 3;
		long one = (r + 2) / 3 - (l + 1) / 3;
		long mod = (long)1e9 + 7;
		dp[1][0] = three; dp[1][1] = one; dp[1][2] = two;
		for (int i = 2; i <= n; i++) {
			dp[i][0] = (dp[i - 1][0] * three % mod + dp[i - 1][1] * two % mod + dp[i - 1][2] * one % mod) % mod;
			dp[i][1] = (dp[i - 1][1] * three % mod + dp[i - 1][2] * two % mod + dp[i - 1][0] * one % mod) % mod;
			dp[i][2] = (dp[i - 1][2] * three % mod + dp[i - 1][0] * two % mod + dp[i - 1][1] * one % mod) % mod;
		}
		System.out.println(dp[n][0]);
	}
}