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 = [int(x) for x in input().split()]
numbers = r-l+1



if l%3 == 0:
    r_0_s = ceil(numbers/3)
    r_1_s = ceil((numbers-1)/3)
    r_2_s = ceil((numbers-2)/3)
elif l%3 == 1:
    r_0_s = ceil((numbers-2)/3)
    r_1_s = ceil((numbers)/3)
    r_2_s = ceil((numbers-1)/3)
else:
    r_0_s = ceil((numbers-1)/3)
    r_1_s = ceil((numbers-2)/3)
    r_2_s = ceil((numbers)/3)
    

r_0 = r_0_s
r_1 = r_1_s
r_2 = r_2_s
for i in range(1,n):
    r_0_temp = (r_0*r_0_s + r_1*r_2_s+r_2*r_1_s) % 1000000007
    r_1_temp = (r_1*r_0_s + r_0*r_1_s +r_2*r_2_s ) % 1000000007
    r_2_temp = (r_1*r_1_s + r_2*r_0_s + r_0*r_2_s) % 1000000007
    r_0 = r_0_temp
    r_1 = r_1_temp
    r_2 = r_2_temp
    #print(r_0,r_1,r_2)
print(r_0 % 1000000007)


