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.*;
public class Abc {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int l = in.nextInt();
		int r = in.nextInt();

		int rZero = (r / 3) - ((l - 1) / 3);
		int total = r - l + 1;
		int rOne = getVal(r) - getVal(l - 1);
		int rTwo = total - rZero - rOne;
		//System.out.println(rZero+" "+rOne+" "+rTwo);

		long mod = 1000000007;
		long dp[][] = new long[n][3];
		for(int i = 0; i < n ; i++) {
			if(i == 0) {
				dp[i][0] = rZero;
				dp[i][1] = rOne;
				dp[i][2] = rTwo; 
			} else {
				dp[i][0] = (dp[i - 1][0] * rZero) % mod + (dp[i - 1][1] * rTwo) % mod + (dp[i - 1][2] * rOne) % mod;
				dp[i][0] %= mod;
				dp[i][1] = (dp[i - 1][0] * rOne) % mod + (dp[i - 1][1] * rZero) % mod + (dp[i - 1][2] * rTwo) % mod;
				dp[i][1] %= mod;
				dp[i][2] = (dp[i - 1][0] * rTwo) % mod + (dp[i - 1][1] * rOne) % mod + (dp[i - 1][2] * rZero) % mod;
				dp[i][2] %= mod;
			}
		}
		System.out.println(dp[n - 1][0]);
	}
	static int getVal(int n) {
		if(n % 3 == 0) {
			return n / 3;
		} else {
			return (int) Math.ceil(n / 3.0);
		}
	}
}