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].
from math import ceil
n, l, r = map(int, input().split())
dp_0 = [0 for i in range(n)]
dp_1 = [0 for i in range(n)]
dp_2 = [0 for i in range(n)]
t = 10**9 + 7
dp_0[0] = (r//3 - ceil(l/3) + 1)
dp_1[0] = ((r - 1)//3 - ceil((l - 1)/3) + 1)
dp_2[0] = ((r - 2)//3 - ceil((l - 2)/3) + 1)

for i in range(1, n):
    dp_0[i] = ((dp_0[i - 1] * dp_0[0])%t + (dp_1[i - 1] * dp_2[0])%t + (dp_2[i - 1] * dp_1[0])%t)%t
    dp_1[i] = ((dp_0[i - 1] * dp_1[0])%t + (dp_1[i - 1] * dp_0[0])%t + (dp_2[i - 1] * dp_2[0])%t)%t
    dp_2[i] = ((dp_0[i - 1] * dp_2[0])%t + (dp_1[i - 1] * dp_1[0])%t + (dp_2[i - 1] * dp_0[0])%t)%t

# print(dp_0)
# print(dp_1)
# print(dp_2)
print(dp_0[n - 1]%t)