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].
n, l, r= map(int, input().split())

mod = 10**9 + 7

nums = ((r - l) // 3 + 1, ((r - l) - 1) // 3 + 1, ((r - l) - 2) // 3 + 1)

if l % 3 == 0:
    n0, n1, n2 = nums
elif l % 3 == 1:
    n1, n2, n0 = nums
else:
    n2, n0, n1 = nums

a0, a1, a2 = n0, n1, n2
for _ in range(1, n):
    aux0 = a0
    aux1 = a1
    aux2 = a2
    a0 = (aux0 * n0 + aux1 * n2 + aux2 * n1) % mod
    a1 = (aux0 * n1 + aux1 * n0 + aux2 * n2) % mod
    a2 = (aux0 * n2 + aux1 * n1 + aux2 * n0) % mod

print(a0)
