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.*;

public class Main {

    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        int n = Integer.parseInt(reader.next());
        int l = Integer.parseInt(reader.next());
        int r = Integer.parseInt(reader.next());
        long[][] dp = new long[n][3];
        int MOD = (int) Math.pow(10,9) + 7;
        int m0 = r / 3 - l / 3;
        if(l % 3 == 0){
            m0 += 1;
        }
        int m1 = (r / 3) - (l % 3 == 0 ? l / 3 : (l % 3 == 1 ? l / 3 : l / 3 + 1));
        if((r % 3 >= 1)){
            m1 += 1;
        }
        int m2 = r - l + 1 - (m0 + m1);
        dp[0][0] = m0;
        dp[0][1] = m1;
        dp[0][2] = m2;
        assert(m0 + m1 + m2 == r - l + 1);
        for(int i = 1; i < n; i++){
            dp[i][0] = (dp[i-1][0]*dp[0][0] + dp[i-1][1]*dp[0][2] + dp[i-1][2]*dp[0][1]) % MOD;
            dp[i][1] = (dp[i-1][0]*dp[0][1] + dp[i-1][1]*dp[0][0] + dp[i-1][2]*dp[0][2]) % MOD;
            dp[i][2] = (dp[i-1][0]*dp[0][2] + dp[i-1][1]*dp[0][1] + dp[i-1][2]*dp[0][0]) % MOD;
        }
        int res = (int) dp[n-1][0] % MOD;
        System.out.println(res);
    }
}
