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.InputStream;
import java.io.PrintStream;
import java.util.function.Supplier;
import java.util.*;

public class Ayoub {
    static Supplier<InputStream> inputSupplier = () -> System.in;
    static Supplier<PrintStream> outSupplier = () -> System.out;
    static int MOD = 1_000_000_007;
    public static void main(String[] args) {
        Scanner sc = new Scanner(inputSupplier.get());
        PrintStream out = outSupplier.get();
        int n = sc.nextInt();
        int l = sc.nextInt();
        int r = sc.nextInt();

        long dp[][] = new long[3][200001];
        dp[0][0] = 1;
        dp[1][0] = 0;
        dp[2][0] = 0;
        int count = r-l+1;
        
        int countMod3[] = new int[3];
        countMod3[0] = count/3;
        countMod3[1] = count/3;
        countMod3[2] = count/3;
        if(count%3==1) {
            countMod3[l%3]++;
        } else if(count%3==2) {
            countMod3[l%3]++;
            countMod3[r%3]++;
        }
        
        for (int i = 1; i <= n; i++) {
            dp[0][i] =
                    ((dp[0][i - 1]*countMod3[0])%MOD +
                    (dp[1][i - 1]*countMod3[2])%MOD +
                    (dp[2][i - 1]*countMod3[1])%MOD)%MOD;

            dp[1][i] =
                    ((dp[0][i - 1]*countMod3[1])%MOD +
                    (dp[1][i - 1]*countMod3[0])%MOD +
                    (dp[2][i - 1]*countMod3[2])%MOD)%MOD;

            dp[2][i] =
                    ((dp[0][i - 1]*countMod3[2])%MOD +
                    (dp[1][i - 1]*countMod3[1])%MOD +
                    (dp[2][i - 1]*countMod3[0])%MOD)%MOD;
        }
        int result = (int)dp[0][n];
        out.println(result);
    }
}
