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
import time
MOD=10**9+7
n,l,r=map(int,raw_input().split())
start=time.time()
dp=[[0]*n for _ in xrange(3)]
k=[0]*3
lm=l%3
rm=r%3
num=r-l+1
if lm==0:
	if rm==0:
		num-=1
		[k[0],k[1],k[2]]=[num/3]*3
		k[0]+=1
	elif rm==1:
		num+=1
		[k[0],k[1],k[2]]=[num/3]*3
		k[2]-=1
	else:
		[k[0],k[1],k[2]]=[num/3]*3
elif lm==1:
	if rm==0:
		[k[0],k[1],k[2]]=[num/3]*3
	elif rm==1:
		num-=1
		[k[0],k[1],k[2]]=[num/3]*3
		k[1]+=1
	else:
		num+=1
		[k[0],k[1],k[2]]=[num/3]*3
		k[0]-=1
else:
	if rm==0:
		num+=1
		[k[0],k[1],k[2]]=[num/3]*3
		k[1]-=1
	elif rm==1:
		[k[0],k[1],k[2]]=[num/3]*3
	else:
		num-=1
		[k[0],k[1],k[2]]=[num/3]*3
		k[2]+=1
dp[0][0]=k[0]
dp[1][0]=k[1]
dp[2][0]=k[2]
for i in xrange(1,n):
	f0=dp[0][i-1]%MOD
	f1=dp[1][i-1]%MOD
	f2=dp[2][i-1]%MOD
	dp[0][i]=(((f0)*(k[0]%MOD))%MOD+((f1)*(k[2]%MOD))%MOD+((f2)*(k[1]%MOD)%MOD))%MOD
	dp[1][i]=(((f0)*(k[1]%MOD))%MOD+((f1)*(k[0]%MOD))%MOD+((f2)*(k[2]%MOD)%MOD))%MOD
	dp[2][i]=(((f0)*(k[2]%MOD))%MOD+((f1)*(k[1]%MOD))%MOD+((f2)*(k[0]%MOD)%MOD))%MOD
print(dp[0][n-1])





