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].
def count(l,r):
	diff=r-l+1
	ans=[diff/3,diff/3,diff/3]
	if diff%3==0:
		return ans
	if diff%3==1:
		ans[l%3]+=1
		return ans
	else:
		ans[l%3]+=1
		ans[((l%3)+1)%3]+=1
		return ans


mod=1000000007
n,l,r=map(int,raw_input().split())
# n,l,r=2,1,10000000
x,y,z=count(l,r)
if n==1:
	print x
	exit()
ans0=[1 for i in range(n+1)]
ans1=[1 for i in range(n+1)]
ans2=[1 for i in range(n+1)]

ans0[0]=1
ans0[1]=x
ans0[2]=(x*x)+(2*y*z)

ans1[0]=1
ans1[1]=y
ans1[2]=(z*z)+(2*x*y)

ans2[0]=1
ans2[1]=z
ans2[2]=(y*y)+(2*x*z)



for i in range(3,n+1):
	ans0[i]=((x*ans0[i-1])%mod+(y*ans2[i-1])%mod+(z*ans1[i-1])%mod)%mod
	ans1[i]=((x*ans1[i-1])%mod+(y*ans0[i-1])%mod+(z*ans2[i-1])%mod)%mod
	ans2[i]=((x*ans2[i-1])%mod+(y*ans1[i-1])%mod+(z*ans0[i-1])%mod)%mod
print ans0[-1]%mod