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 ayoubandlostarray;

import java.util.*;
import java.io.*;

public class ayoubandlostarray {
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		
		long[] op = new long[3];	//stores how many different numbers there are % 3
		StringTokenizer st = new StringTokenizer(fin.readLine());
		int n = Integer.parseInt(st.nextToken());
		int l = Integer.parseInt(st.nextToken());
		int r = Integer.parseInt(st.nextToken());
		int mod = 1000000007;
		long[][] dp = new long[n][3];
		int base = (r - l + 1) / 3;
		int offset = l % 3;	//huh
		int remainder = (r - l + 1) % 3;
		Arrays.fill(op, base);
		for(int i = offset; i < offset + remainder; i++) {
			op[i % 3] ++;
		}
		dp[0][0] = op[0];
		dp[0][1] = op[1];
		dp[0][2] = op[2];
		
		for(int i = 1; i < n; i++) {
			
			for(int j = 0; j < 3; j++) {
				for(int k = 0; k < 3; k++) {
					dp[i][(j + k) % 3] += dp[i - 1][j] * op[k];
					dp[i][(j + k) % 3] %= mod;
				}
			}
//			for(long j : dp[i]) {
//				System.out.print(j + " ");
//			}
//			System.out.println();
		}
		System.out.println(dp[n - 1][0]);
	}
}
