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 math


modby = (10 ** 9 + 7)


if __name__ == '__main__':
    n, l, r = list(map(int, input().split()))
    mod_counts = [
        math.floor(r/3) - math.ceil(l/3) + 1, math.floor((r-1)/3) - math.ceil((l-1)/3) + 1,
        math.floor((r+1)/3) - math.ceil((l+1)/3) + 1]
    prev_0, prev_1, prev_2 = mod_counts
    cur_0, cur_1, cur_2 = 0, 0, 0
    for _ in range(1, n):
        cur_0 = ((prev_0 * mod_counts[0]) + (prev_1 * mod_counts[2]) + (prev_2 * mod_counts[1])) % modby
        cur_1 = ((prev_0 * mod_counts[1]) + (prev_1 * mod_counts[0]) + (prev_2 * mod_counts[2])) % modby
        cur_2 = ((prev_0 * mod_counts[2]) + (prev_1 * mod_counts[1]) + (prev_2 * mod_counts[0])) % modby
        prev_0, prev_1, prev_2 = cur_0, cur_1, cur_2
    print(prev_0)
