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.*;
import java.io.*;

public class C {
    static final long MOD = 1000000007;
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int N = scanner.nextInt();
        int L = scanner.nextInt();
        int R = scanner.nextInt();
        int tempL = L;
        int tempR = R;
        while(tempL % 3 != 0) tempL++;
        while(tempR % 3 != 0 && R - tempR < 3) tempR--;
        long nModz = 1 + (tempR-tempL)/3;
        tempL = L;
        tempR = R;
        while(tempL % 3 != 1) tempL++;
        while(tempR % 3 != 1 && R - tempR < 3) tempR--;
        long nModo = 1 + (tempR-tempL)/3;
        tempL = L;
        tempR = R;
        while(tempL %3 != 2) tempL++;
        while(tempR %3 != 2 && R - tempR < 3) tempR--;
        long nModt = 1 + (tempR - tempL)/3;
        long prevZ = nModz;
        long prevO = nModo;
        long prevT = nModt;
        for(int i = 1; i < N; i++) {
            long newO, newT, newZ;
            newZ = (prevT * nModo + prevO * nModt + prevZ * nModz) %MOD;
            newO = (prevT * nModt + prevZ * nModo + prevO * nModz) %MOD;
            newT = (prevZ * nModt + prevO * nModo + prevT * nModz) %MOD;
            prevZ = newZ;
            prevO = newO;
            prevT = newT;
        }
        out.println(prevZ);
        out.flush();
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        public FastScanner() {
            this(new InputStreamReader(System.in));
        }

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

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }
}
