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].
nlr=list(map(int,input().split()))
n=nlr[0]
l=nlr[1]
r=nlr[2]
p=10**9+7
L=[[1]*(3) for j in range(n+1)]
M=[0]*3
z=r-l+1
if(z==1):
	if(l%3==0):
		print(1)
	else:
		if(n%3==0):
			print(1)
		else:
			print(0)
else:
	if(z%3==0):
		M[0]=M[1]=M[2]=z//3
	else:
		if(z%3==1):
			M[0]=z//3
			M[1]=z//3
			M[2]=z//3
			M[l%3]+=1
		else:
			M[0]=1+z//3
			M[1]=1+z//3
			M[2]=1+z//3
			M[(r+1)%3]-=1

	
	for i in range(1,n+1):
		if(i==1):
			L[i][0]=M[0]
			L[i][1]=M[1]
			L[i][2]=M[2]
		else:
			L[i][0]=(L[i-1][0]*M[0]+L[i-1][1]*M[2]+L[i-1][2]*M[1])%p
			L[i][1]=(L[i-1][1]*M[0]+L[i-1][0]*M[1]+L[i-1][2]*M[2])%p
			L[i][2]=(L[i-1][2]*M[0]+L[i-1][1]*M[1]+L[i-1][0]*M[2])%p

	print(L[n][0])


