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].
//package ;
import java.io.*;
import java.util.*;

public class D {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);
		n=sc.nextInt();
		pos=new long[3];
		dp=new long[n][3];
		for(int i=0;i<n;i++)Arrays.fill(dp[i], -1);
		long l=sc.nextLong(),r=sc.nextLong();
		long x=(r/3)-((l-1)/3);//3add el mult of 3
		pos[0]=x;
		pos[1]=x-1;
		pos[2]=x-1;
		if(r%3==1)
			pos[1]++;
		if(r%3==2)
		{
			pos[1]++;
			pos[2]++;
		}
		if(l%3==2)
			pos[2]++;
		if(l%3==1)
		{
			pos[1]++;
			pos[2]++;
		}
//		System.out.println(Arrays.toString(pos));
		pw.print(dp(0,0));
		pw.close();
	}
	static long mod=(long)1e9+7;
	static int n;
	static long dp[][];
	static long pos[];
	static long dp(int i,int sum)
	{
		if(i==n)return sum==0?1:0;
		if(dp[i][sum]!=-1)return dp[i][sum];
		long ans=0;
		ans=ans+(dp(i+1,(sum+1)%3)*pos[1])%mod;
		ans%=mod;
		ans=ans+(dp(i+1,(sum+2)%3)*pos[2])%mod;
		ans%=mod;
		ans=ans+(dp(i+1,sum)*pos[0])%mod;
		ans%=mod;
		return dp[i][sum]=ans;
	}
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		boolean hasnext() throws IOException {
			return br.ready();
		}

	}
}
