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].
n, l, r = map(int, raw_input().split()) 

mod = 1000000007

aux = [0] * 3
a = [0] * (n+1) 
b = [0] * (n+1)
c = [0] * (n+1)

aux[0] = aux[1] = aux[2] = (r-l+1)/3

if (r-l+1) % 3 == 2:
  aux[1] += 1 
  aux[2] += 1 
elif (r-l+1) % 3 == 1:
  aux[1] += 1 

if l%3 == 2:
  aux = [aux[2], aux[0], aux[1]]
elif l%3==0:
    aux=[aux[1],aux[2],aux[0]] 

c[1] = aux[2]
a[1] = aux[0]
b[1] = aux[1]

for i in range(2, n+1):
  
  a[i]=(a[i-1]*aux[0] + b[i-1]*aux[2] + c[i-1]*aux[1]) % mod
    
  b[i]=(a[i-1]*aux[1] + b[i-1]*aux[0] + c[i-1]*aux[2]) % mod
  
  c[i]=(a[i-1]*aux[2]+ b[i-1]*aux[1] + c[i-1]*aux[0]) % mod 

print a[n] % mod
   	   		    	   					 	 		 		