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 
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
mod = int(1e9)+7
def ip(mode):
	if mode==0:
		return int(input())
	if mode==1:
		return list(map(int, input().split()))
	if mode==2:
		return input()
	if mode==3:
		return list(map(str, input().split()))
	if mode==4:
		return map(int, input().split())
	if mode==5:
		return map(str, input().split())

for _ in range(1):
	n,l,r = ip(4)
	dp = [[0,0,0] for i in range(n)]
	m0 = (r//3)  - (l//3 + (l%3!=0)) + 1
	m1 = ((r-1)//3)  - ((l-1)//3 + ((l-1)%3!=0)) + 1
	m2 = ((r-2)//3)  - ((l-2)//3 + ((l-2)%3!=0)) + 1
	# print(m0,m1,m2)
	dp[0][0] = m0
	dp[0][1] = m1
	dp[0][2] = m2
	for i in range(1,n):
		dp[i][0] = ((dp[i-1][0]*m0) + (dp[i-1][1]*m2) + (dp[i-1][2]*m1))%mod
		dp[i][1] = ((dp[i-1][0]*m1) + (dp[i-1][1]*m0) + (dp[i-1][2]*m2))%mod
		dp[i][2] = ((dp[i-1][0]*m2) + (dp[i-1][1]*m1) + (dp[i-1][2]*m0))%mod
	print(dp[n-1][0]%mod)