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].
/*
https://codeforces.com/problemset/problem/1105/C

divisible by 3:
    numWays(n, 0):
        (x % 3)  numWays(n - 1, (3 - (x % 3)) % 3)

    O(n * (r - l + 1))
    
    
  x -> [l, r] : x % 3 -> 0, 1, 2
  
  calculate by group
  [l, r ] = > 0 .... m
              1 .... q
              2 .... k
*/
import java.util.Scanner;
public class AyoubAndLostArray {
    private static int MOD = 1000000007;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int l = sc.nextInt();
        int r = sc.nextInt();
        // count number of number that modules 3 equals 0, 1, 2.
        int[] cnt = new int[3];
        for (int i = 0; i < 3; i++) {
            cnt[(l + i) % 3] = (r - l - i) >= 0 ? ((r - l - i) / 3 + 1) : 0; 
        }

        long[][] L = new long[n][3];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 3; j++) {
                if (i == 0) {
                    L[i][j] = cnt[j];
                    continue;
                }
                for (int k = 0; k < 3; k++) {
                    L[i][j] = (L[i][j] + cnt[k] * L[i - 1][(j + 3 - k) % 3] % MOD) % MOD;
                }
                // if (j != 0) {
                //     L[i][j] = (L[i][j] + (cnt[j] * L[i - 1][0]) % MOD) % MOD;
                //     L[i][j] = (L[i][j] + (cnt[0] * L[i - 1][j]) % MOD) % MOD;
                //     L[i][j] = (L[i][j] + (cnt[3 - j] * L[i - 1][3 - j]) % MOD) % MOD;
                // }
                // else {
                //     L[i][j] = (L[i][j] + (cnt[j] * L[i - 1][0]) % MOD) % MOD;
                //     L[i][j] = (L[i][j] + (cnt[2] * L[i - 1][1]) % MOD) % MOD;
                //     L[i][j] = (L[i][j] + (cnt[1] * L[i - 1][2]) % MOD) % MOD;
                // }
            }
        }

        System.out.println(L[n-1][0]);
    }
}
