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())
x, y, z = 0, 0, 0
i = l
j = r
mod = 10**9 + 7 
if((l % 3) == 1):
  x += 1
  y += 1
  i = (l + 2)
if((r % 3) == 1):
  x += 1
  j = (r - 1)
if((l % 3) == 2):
  y += 1
  i = (l + 1)
if((r % 3) == 2):
  x += 1
  y += 1
  j = (r - 2)
 
w = ((j - i) // 3)
i = x = (x + w)
j = y = (y + w)
res = z = (z + w + 1)
 
for _ in range(n - 1):
    aux_1 = (i * z) + (j * y) + (res * x)
    aux_2 = (i * x) + (j * z) + (res * y)
    aux_3 = (i * y) + (j * x) + (res * z)
    
    i = aux_1 % mod
    j = aux_2 % mod
    res = aux_3 % mod
 
print(res)
