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.*;

import static java.lang.Math.*;

public class Main {
    void problem() {
        int n = in.nextInt();
        int l = in.nextInt();
        int r = in.nextInt();

        int cur = l;
        long n1 = 0;
        long n2 = 0;
        long n0 = 0;
        for (int i = 0; i < 3 && cur <= r; i++) {
            switch (cur % 3) {
                case (0):
                    n0 = 1 + (r - cur) / 3;
                    break;
                case (1):
                    n1 = 1 + (r - cur) / 3;
                    break;
                case (2):
                    n2 = 1 + (r - cur) / 3;
                    break;
            }
            cur++;
        }

        long dp[][] = new long[n][3];
        dp[0][0] = n0;
        dp[0][1] = n1;
        dp[0][2] = n2;
        for (int i = 1; i < n; i++) {
            dp[i][0] = (dp[i - 1][0] * n0) % N;
            dp[i][0] = (dp[i][0] + dp[i - 1][1] * n2) % N;
            dp[i][0] = (dp[i][0] + dp[i - 1][2] * n1) % N;

            dp[i][1] = (dp[i - 1][0] * n1) % N;
            dp[i][1] = (dp[i][1] + dp[i - 1][1] * n0) % N;
            dp[i][1] = (dp[i][1] + dp[i - 1][2] * n2) % N;

            dp[i][2] = (dp[i - 1][0] * n2) % N;
            dp[i][2] = (dp[i][2] + dp[i - 1][1] * n1) % N;
            dp[i][2] = (dp[i][2] + dp[i - 1][2] * n0) % N;
        }
        out.println(dp[n-1][0]);
    }

    FastScanner in;
    PrintWriter out;

    static long N = (int)pow(10, 9) + 7;

    void run() {
        in = new FastScanner();
        out = new PrintWriter(System.out);
        problem();
        out.close();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastScanner(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }

        public long nextLong() {
            return Long.parseLong(nextToken());
        }

        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }

    public static void main(String[] args) {
        new Main().run();
    }

    public static long modPower(int x, int y) {
        if(y==0) return 1;
        long z = modPower(x, y/2);
        if ((y % 2) == 0)
            return (z*z)%N;
        else
            return (x*z*z)%N;
    }
}