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.*;

public class Main {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(in, out);
        out.close();
    }

    public static class TaskC {
        public void solve(InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int l = in.nextInt();
            int r = in.nextInt();
            long m0 = amountInSegment(l, r, 0);
            long m1 = amountInSegment(l, r, 1);
            long m2 = amountInSegment(l, r, 2);
            long dp[][] = new long[n + 1][3];
            dp[1][0] = m0;
            dp[1][1] = m1;
            dp[1][2] = m2;
            long MOD = 1000000007;
            for (int i = 2; i <= n; i++) {
                dp[i][0] = dp[i - 1][0] * m0 + dp[i - 1][1] * m2 + dp[i - 1][2] * m1;
                dp[i][1] = dp[i - 1][0] * m1 + dp[i - 1][1] * m0 + dp[i - 1][2] * m2;
                dp[i][2] = dp[i - 1][0] * m2 + dp[i - 1][2] * m0 + dp[i - 1][1] * m1;
                dp[i][0] %= MOD;
                dp[i][1] %= MOD;
                dp[i][2] %= MOD;
            }
            out.println(dp[n][0]);
        }

        private long amountInSegment(int l, int r, int mod) {
            double left = Math.ceil((l - mod) / 3.0);
            double right = Math.floor((r - mod) / 3.0);
            return (long) (right - left + 1);
        }
    }

    public static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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