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.*;
import java.io.*;
import static java.lang.Math.*;

public class Solution {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n, l, r;
		int MOD = 1000000007;
		n = sc.nextInt();
		l = sc.nextInt();
		r = sc.nextInt();
		int k[] = { 0, 0, 0 };
		int g = (r - l + 1) / 3;
		k[0] = k[1] = k[2] = g;
		if ((r - l + 1) % 3 == 1) {
			k[l % 3]++;
		} else if ((r - l + 1) % 3 == 2) {
			k[r % 3]++;
			k[(r - 1) % 3]++;
		}
		long dp[][] = new long[n + 1][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]) % MOD;
			dp[i][1] = (dp[i - 1][0] * k[1] + dp[i - 1][1] * k[0] + dp[i - 1][2] * k[2]) % MOD;
			dp[i][2] = (dp[i - 1][0] * k[2] + dp[i - 1][1] * k[1] + dp[i - 1][2] * k[0]) % MOD;
		}

		System.out.println(dp[n][0]);
	}
}
