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
mod = 10 ** 9 + 7

def f(l, r, x):
	return ceil((r - x) / 3) - ceil((l - 1 - x) / 3)

n, l, r =map(int, input().split())
m1 = f(l, r, 0)
m2 = f(l, r, 1)
m0 = f(l, r ,2)
dp0 = [1 for i in range(n+1)]
dp1 = [0 for i in range(n+1)]
dp2 = [0 for i in range(n+1)]
for i in range(1, n+1):
	dp0[i] = (dp0[i-1] * m0 + dp1[i-1] * m2 + dp2[i-1] * m1)%mod
	dp1[i] = (dp1[i-1] * m0 + dp2[i-1] * m2 + dp0[i-1] * m1)%mod
	dp2[i] = (dp2[i-1] * m0 + dp0[i-1] * m2 + dp1[i-1] * m1)%mod
print(dp0[n])