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 = list(map(int,input().split()))

A = [[0,0,0]]
A[0][0],A[0][1],A[0][2] = (r-l+1)//3,(r-l+1)//3,(r-l+1)//3

if (r - l + 1) % 3 == 1:
    A[0][(l)%3] += 1
elif (r - l + 1) % 3 == 2:
    A[0][r%3] += 1
    A[0][l%3] += 1

for i in range(1,n):
    z = (A[-1][0] * A[0][0])%(10**9+7) + (A[-1][1] * A[0][2])%(10**9+7) + (A[-1][2] * A[0][1])%(10**9+7)
    o = (A[-1][0] * A[0][1])%(10**9+7) + (A[-1][1] * A[0][0])%(10**9+7) + (A[-1][2] * A[0][2])%(10**9+7)  
    t = (A[-1][0] * A[0][2])%(10**9+7) + (A[-1][1] * A[0][1])%(10**9+7) + (A[-1][2] * A[0][0])%(10**9+7)
    
    A.append([z%(10**9+7),o%(10**9+7),t%(10**9+7)])

print(A[-1][0]%(10**9+7))
