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.*;
import java.util.stream.*;

@SuppressWarnings("unused")
public class AyoubAndLostArray {
	public static long MOD = (long) (1e9+7);
	public static void main(String[] args) throws IOException {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), l = in.nextInt(), r = in.nextInt();
		int num0, num1, num2;
		num0 = num1 = num2 = (r-l+1)/3;
		switch(3* (l%3)+(r%3)) {
			case 1: num1++;
			case 0: num0++;
			case 2:	break;
			case 5: num2++;
			case 4: num1++;
			case 3: break;
			case 6: num0++;
			case 8: num2++;
			case 7: break;
		}
		long[][] dp = new long[n+1][3];
		dp[0][0] = 1;
		for (int i = 1; i <= n; i++) {
			dp[i][0] = mod(  mod(dp[i-1][0]*num0) + mod(dp[i-1][1]*num2) + mod(dp[i-1][2]*num1)  );
			dp[i][1] = mod(  mod(dp[i-1][0]*num1) + mod(dp[i-1][1]*num0) + mod(dp[i-1][2]*num2)  );
			dp[i][2] = mod(  mod(dp[i-1][0]*num2) + mod(dp[i-1][1]*num1) + mod(dp[i-1][2]*num0)  );
		}
		System.out.println(dp[n][0]);
	}
	public static long mod(long l) {
		return l % MOD;
	}
}
