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 math import ceil
n,a,b=list(map(int,input().split()))
d=[0]*3 
s=b-a+1
mod=10**9+7
t=s%3
x,y,z=[ceil(s/3)]*t+[s//3]*(3-t)
if a%3==1:
    y,z,x=x,y,z
elif a%3==2:
    z,x,y=x,y,z
d[0],d[1],d[2]=x,y,z
a1,a2,a3=x,y,z
for i in range(1,n):
    d[0]=((a1*x)+(a2*z)+(a3*y))%mod
    d[1]=((a1*y)+(a2*x)+(a3*z))%mod
    d[2]=((a1*z)+(a2*y)+(a3*x))%mod
    a1,a2,a3=d[0],d[1],d[2]
print(d[0]%mod)
