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].
inf = pow(10,9)+7


n,l,r = map(int,input().split())
list1 = [[],[],[]]


for i in range(n):
    list1[0].append(0)
    list1[1].append(0)
    list1[2].append(0)
list2 = list()

s = l%3
x,y,z = 0,0,0
if s==0:
    x = (r-l)//3 + 1
    y = (r-l-1)//3 +1
    z = (r-l-2)//3+1
elif s==1:
    y= (r-l)//3 +1
    z = (r-1-l)//3+1
    x = (r-l-2)//3 +1
else:
    z = (r-l)//3+1
    x = (r-l-1)//3 +1
    y = (r-l-2)//3 +1

list1[0][0]=x%inf
list1[1][0]=y%inf
list1[2][0]=z%inf



for i in range(1,n):

    zeroth = (list1[0][0]*list1[0][i-1]) %inf+ (list1[1][i-1]*list1[2][0])%inf + (list1[2][i-1]*list1[1][0])%inf
    one = (list1[0][0]*list1[1][i-1])%inf + (list1[1][0]*list1[0][i-1])%inf + (list1[2][i-1]*list1[2][0])%inf
    two = (list1[0][i-1]*list1[2][0])%inf + (list1[2][i-1]*list1[0][0])%inf + (list1[1][i-1]*list1[1][0])%inf
    list1[0][i]+=zeroth%inf
    list1[1][i]+=one%inf
    list1[2][i]+=two%inf

print(list1[0][n-1]%inf)
