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.io.*;
import java.util.ArrayList;
import java.util.List;

public class Main {

    private static final boolean FROM_FILE = false;
    private static final boolean TO_FILE = false;

    public static void main(String[] args) throws IOException {
        BufferedReader br;
        if (FROM_FILE) {
            br = new BufferedReader(new FileReader("C:/home/programming/input.txt"));
        } else {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        PrintWriter pw;
        if (TO_FILE) {
            pw = new PrintWriter("C:/home/programming/google/A_large_output.txt");
        } else {
            pw = new PrintWriter(System.out);
        }

        // Algorithm:

        String[] line = br.readLine().split(" ");

        int n = Integer.parseInt(line[0]);
        int l = Integer.parseInt(line[1]);
        int r = Integer.parseInt(line[2]);

        pw.println(calculate(n, l, r));

        pw.flush();
    }

    private static int calculate(int n, int l, int r) {
        int[][] cache = new int[n + 1][3];
        for (int i = 0; i < n + 1; i++) {
            for (int j = 0; j < 3; j++) {
                cache[i][j] = -1;
            }
        }
        return f(n, l, r, 0, cache);
    }

    private static int f(int n, int l, int r, int flag, int[][] cache) {
        if (cache[n][flag] != -1) {
            return cache[n][flag];
        }
        if (n == 0) {
            return flag == 0 ? 1 : 0;
        }
        long total = 0;
        for (int i = l; i <= Math.min(l + 2, r); i++) {
            int rest = i % 3;
            total += ((((r - i) / 3) + 1L) * f(n - 1, l, r, (rest + flag) % 3, cache)) % 1_000_000_007;
            total = total % 1_000_000_007;
        }
        cache[n][flag] = (int) total;
        return (int) total;
    }
}