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.lang.*;
import java.io.*;


public class Main {
	static int N,L,R;
	static int mod = 1000000007;
	static long [][] dp = new long[200001][3];
	static long [] a = new long[3];

	static long add(long a, long b) {
		a += b;
		if(a >= mod) return a - mod;
		return a;
	}

	static long mul(long a, long b) {
		a *= b;
		if(a >= mod) return a % mod;
		return a;
	}


	static long numberOfArray(int i,int rem){
		if (i==0) {
			if(rem==0) return 1;
			else return 0;
		}
		else if (dp[i][rem]!=-1) {
			return dp[i][rem];
		}
		else{
			long res = 0;
			res = add(res,mul(a[0],numberOfArray(i-1,((rem + 3 ) % 3))));
			res = add(res,mul(a[1],numberOfArray(i-1,((rem + 3 - 1) % 3))));
			res = add(res,mul(a[2],numberOfArray(i-1,((rem + 3 - 2) % 3))));
			return dp[i][rem] = res;
		}
		
	}

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		N = scan.nextInt(); L = scan.nextInt() ; R = scan.nextInt();
		int rem_0 = -1,rem_1 = -1,rem_2 = -1;
			int r_l = L%3,r_r = R%3;
			if (r_l==0 && r_r==0) {
				rem_0 = ((R-L)/3)+1;
				rem_1 = ((R-L)/3);
				rem_2 = ((R-L)/3);	
			}
			else if (r_l==0 && r_r==1) {
				rem_0 = ((R-L)/3)+1;
				rem_1 = ((R-L)/3)+1;
				rem_2 = ((R-L)/3);	
			}
			else if (r_l==0 && r_r==2) {
				rem_0 = ((R-L)/3)+1;
				rem_1 = ((R-L)/3)+1;
				rem_2 = ((R-L)/3)+1;
			}
			else if (r_l==1 && r_r==0) {
				rem_0 = ((R-L)/3)+1;
				rem_1 = ((R-L)/3)+1;
				rem_2 = ((R-L)/3)+1;
			}
			else if (r_l==1 && r_r==1) {
				rem_0 = ((R-L)/3);
				rem_1 = ((R-L)/3)+1;
				rem_2 = ((R-L)/3);
			}
			else if (r_l==1 && r_r==2) {
				rem_0 = ((R-L)/3);
				rem_1 = ((R-L)/3)+1;
				rem_2 = ((R-L)/3)+1;
			}
			else if (r_l==2 && r_r==0) {
				rem_0 = ((R-L)/3)+1;
				rem_1 = ((R-L)/3);
				rem_2 = ((R-L)/3)+1;
			}
			else if (r_l==2 && r_r==1) {
				rem_0 = ((R-L)/3)+1;
				rem_1 = ((R-L)/3)+1;
				rem_2 = ((R-L)/3)+1;
			}
			else if (r_l==2 && r_r==2) {
				rem_0 = ((R-L)/3);
				rem_1 = ((R-L)/3);
				rem_2 = ((R-L)/3)+1;
			}
			a[0] = rem_0;a[1] = rem_1;a[2] = rem_2;
		for (int i=0;i<200001 ;i++ ) {
			for (int j=0;j<3 ;j++ ) {
				dp[i][j] = -1;
			}
		}
		
		System.out.println(numberOfArray(N,0));
	}
}
