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,floor

n,l,r = map(int,(input().split(" ")))
mod = 10**9+7

dp0 = [0]*(n)
dp1 = [0]*(n)
dp2 = [0]*(n)



nr0 = floor(r/3)- ceil(l/3) + 1
nr1 = floor((r-1)/3) - ceil((l-1)/3)+1
nr2 = floor((r-2)/3) - ceil((l-2)/3) +1





dp0[0] = nr0
dp1[0] = nr1
dp2[0] = nr2

#print(nr0,nr1,nr2)

for i in range (1,n) :
    dp0[i] += dp0[i-1]*nr0 + dp1[i-1]*nr2 + dp2[i-1]*nr1
    dp0[i] %= mod

    dp1[i] += dp0[i - 1] * nr1 + dp1[i - 1] * nr0 + dp2[i - 1] * nr2
    dp1[i] %= mod

    dp2[i] += dp0[i - 1] * nr2 + dp1[i - 1] * nr1 + dp2[i - 1] * nr0
    dp2[i] %= mod

#for i in range (1,n+1) :
print(dp0[n-1])











