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].
MOD = 10**9+7
import math
def C():

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

	c0 = math.floor(r/3) - math.ceil(l/3) + 1
	c1 = math.floor((r-1)/3) - math.ceil((l-1)/3) + 1
	c2 = math.floor((r-2)/3) - math.ceil((l-2)/3) + 1

	dp = [[0 for _ in range(3)] for __ in range(n+1)]
	dp[1] = [c0%MOD , c1%MOD , c2%MOD]

	for i in range(2,n+1):
		for j in range(3):

			if(j%3 == 0): dp[i][j] = (c0*dp[i-1][0] + c1*dp[i-1][2] + c2*dp[i-1][1])%MOD
			elif(j%3 == 1): dp[i][j] = (c0*dp[i-1][1] + c1*dp[i-1][0] + c2*dp[i-1][2])%MOD
			else: dp[i][j] = dp[i][j] = (c0*dp[i-1][2] + c1*dp[i-1][1] + c2*dp[i-1][0])%MOD

	print(dp[n][0])














C()