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].
tmp = [int(x) for x in input().strip().split(' ')]
M = 1000000007

size = tmp[0]
l = tmp[1]
r = tmp[2]

mod = [0, 0, 0]

while l % 3 != 0 and l < r:
	mod[l % 3] += 1
	l += 1

while r % 3 != 2 and r > l:
	mod[r%3] += 1
	r -= 1

if l == r:
	mod[l%3] += 1
else:
	for i in range(3):
		mod[i] += int((r+1-l) / 3)


cur_res = [mod[0], mod[1], mod[2]]


for i in range(size - 1):
	next_res = [0, 0, 0]
	next_res[0] = (cur_res[0] * mod[0] + cur_res[1] * mod[2] + cur_res[2] * mod[1]) % M
	next_res[1] = (cur_res[1] * mod[0] + cur_res[2] * mod[2] + cur_res[0] * mod[1]) % M
	next_res[2] = (cur_res[2] * mod[0] + cur_res[0] * mod[2] + cur_res[1] * mod[1]) % M
	cur_res = next_res

print(cur_res[0])
