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.Scanner;

public class Main{
    private static final long MODULA = 1_000_000_007;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int l = scanner.nextInt();
        int r = scanner.nextInt();
        long[][] dp = new long[3][n + 1];
        long[] c = new long[3];

        c[0] = kinch(l, r, 0);
        c[1] = kinch(l, r, 1);
        c[2] = kinch(l, r, 2);

        dp[0][1] = c[0];
        dp[1][1] = c[1];
        dp[2][1] = c[2];

        for (int i = 2 ; i <= n; ++i ) {
            dp[0][i] = ((dp[1][i - 1] * c[2] ) % MODULA + (dp[2][i - 1] * c[1]) % MODULA + (dp[0][i - 1]* c[0]) % MODULA + MODULA) % MODULA;
            dp[1][i] = ((dp[0][i - 1] * c[1]) % MODULA + (dp[2][i - 1] * c[2]) % MODULA + (dp[1][i - 1] * c[0]) % MODULA + MODULA) % MODULA;
            dp[2][i] = ((dp[2][i - 1] * c[0]) % MODULA + (dp[1][i - 1] * c[1]) % MODULA + (dp[0][i - 1] * c[2]) % MODULA + MODULA) % MODULA;
        }
        System.out.println(dp[0][n]);
    }

    private static int kinch (int l, int r, int c) {
        while (l % 3 != c) ++l;
        while ((r + 3) % 3 != c) --r;
        if (l > r) return 0;
        return (r - l) / 3 + 1;
    }

}