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].
size, left, right = map(int, raw_input().split())
MOD = 10 ** 9 + 7

rest = {
    0: (right / 3) - ((left - 1) / 3),
    1: ((right - 1) / 3) - ((left - 2) / 3),
    2: ((right - 2) / 3) - ((left - 3) / 3)
}

aux = [rest[0], rest[1], rest[2]]

for i in range(2, size + 1):
    aux_0 = (aux[0] * rest[0]) + (aux[1] * rest[2]) + (aux[2] * rest[1])
    aux_1 = (aux[2] * rest[2]) + (aux[0] * rest[1]) + (aux[1] * rest[0])
    aux_2 = (aux[1] * rest[1]) + (aux[0] * rest[2]) + (aux[2] * rest[0])

    aux[0] = aux_0 % MOD
    aux[1] = aux_1 % MOD
    aux[2] = aux_2 % MOD

print(aux[0])
