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 collections import *
from math import *
n,l,r=map(int,input().split(' '))
d=[[0 for i in range(3)] for i in range(n+1)]
d[0][0]=1
a=floor(r/3)-ceil(l/3)+1
b=floor((r-1)/3)-ceil((l-1)/3)+1
c=floor((r-2)/3)-ceil((l-2)/3)+1
mod=10**9+7
for i in range(1,n+1):
    d[i][0]=((((d[i-1][0]*a)+(d[i-1][1]*c))+(d[i-1][2]*b)))%mod
    d[i][1]=((((d[i-1][0]*b)+(d[i-1][1]*a))+(d[i-1][2]*c)))%mod
    d[i][2]=((((d[i-1][0]*c)+(d[i-1][1]*b))+(d[i-1][2]*a)))%mod

print(d[n][0])
