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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author DHV
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CAyoubAndLostArray solver = new CAyoubAndLostArray();
        solver.solve(1, in, out);
        out.close();
    }

    static class CAyoubAndLostArray {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            long MOD = (long) (1e9 + 7);
            int n = in.nextInt();
            int l = in.nextInt();
            int r = in.nextInt();
            int no = r - l + 1;
            long[][] f = new long[n][3];
            if (no % 3 == 0) {
                for (int i = 0; i < 3; i++) {
                    f[0][i] += no / 3;
                }
            } else if (no % 3 == 1) {
                f[0][r % 3] = no / 3 + 1;
                f[0][(r % 3 + 1) % 3] = no / 3;
                f[0][(r % 3 + 2) % 3] = no / 3;
            } else {
                f[0][r % 3] = no / 3 + 1;
                f[0][(r % 3 + 1) % 3] = no / 3;
                f[0][(r % 3 + 2) % 3] = no / 3 + 1;
            }

            for (int i = 1; i < n; i++) {
                f[i][0] += f[i - 1][0] * f[0][0] % MOD;
                f[i][0] += f[i - 1][1] * f[0][2] % MOD;
                f[i][0] += f[i - 1][2] * f[0][1] % MOD;
                f[i][0] %= MOD;
                f[i][1] += f[i - 1][0] * f[0][1] % MOD;
                f[i][1] += f[i - 1][1] * f[0][0] % MOD;
                f[i][1] += f[i - 1][2] * f[0][2] % MOD;
                f[i][1] %= MOD;
                f[i][2] += f[i - 1][0] * f[0][2] % MOD;
                f[i][2] += f[i - 1][1] * f[0][1] % MOD;
                f[i][2] += f[i - 1][2] * f[0][0] % MOD;
                f[i][2] %= MOD;
            }
            out.println(f[n - 1][0]);
        }

    }
}

