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].
from sys import stdin,stdout
from collections import defaultdict
import math
#stdin = open('input.txt', 'r')  
I = stdin.readline
P = stdout.write
	

n,l,r = map(int,I().split())
a,b,c = 0,0,0
if(l%3==0):
	a = ((r-l)//3) + 1
	b =((r-l-1)//3)+1
	c = (r-l+1)-a-b
elif(l%3==1):
	b = ((r-l)//3)+1
	a = ((r-l-2)//3)+1
	c = (r-l+1)-a-b
else:
	c = ((r-l)//3)+1
	a = ((r-l-1)//3)+1
	b = (r-l+1)-a-c
mod = 10**9+7
#print(a,b,c)
dp = [[0 for i in range(3)] for i in range(n+1)]
mul = [a,b,c]
dp[1][0] = a
dp[1][1] = b
dp[1][2] = c
for i in range(2,n+1):
	dp[i][0] = (dp[i-1][0]*a+dp[i-1][1]*c+dp[i-1][2]*b)%mod
	dp[i][1] = (dp[i-1][0]*b+dp[i-1][1]*(a)+dp[i-1][2]*c)%mod
	dp[i][2] = (dp[i-1][0]*c+dp[i-1][1]*b+dp[i-1][2]*a)%mod
	
#print(dp) 
print(dp[n][0])





	

			







      
   






















        
       
    




