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].
import sys
input = sys.stdin.readline

def One(type):
	return (type(input()))
def List(type):
	return list(map(type,input().split()))
def Str():
	return (list(input()))
def Many(type):
	return map(type,input().split())

def m(a):
	return a%1000000007

def main():
	n,l,r = Many(int)
	dp = [[0]*n,[0]*n,[0]*n]

	m0 = r//3-(l-1)//3
	m1 = m0+(1 if l%3==1 else 0)-(1 if r%3==0 else 0)
	m2 = (r-l+1)-(m0+m1)

	dp[0][0] = m0
	dp[1][0] = m1
	dp[2][0] = m2

	for i in range(n-1):
		dp[0][i+1]+=m(dp[1][i]*m2)+m(dp[2][i]*m1)+m(dp[0][i]*m0)
		dp[1][i+1]+=m(dp[0][i]*m1)+m(dp[1][i]*m0)+m(dp[2][i]*m2)
		dp[2][i+1]+=m(dp[0][i]*m2)+m(dp[1][i]*m1)+m(dp[2][i]*m0)

	print(dp[0][n-1]%1000000007)

main()
