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,input().split())
mod = 1000000007
valor1 = 0
valor2 = 0
valor3 = 0
cont1 = l
cont2 = r

if(r % 3 == 1):
  valor1 += 1
  cont2 = r-1
  
if(r % 3 == 2): 
  valor1+=1
  valor2+=1
  cont2 = r-2

if(l % 3 == 1):
  valor1 += 1
  valor2 += 1
  cont1 = l + 2

if(l % 3 == 2): 
  valor2 += 1
  cont1 = l + 1
  
item = (cont2-cont1) // 3

valor1 = valor1 + item
valor2 = valor2 + item
valor3 = valor3 + item + 1
cont1 = valor1
cont2 = valor2
saida = valor3

for num in range(1, n):
    i = ((cont1 * valor3) + (cont2 * valor2) + (saida * valor1)) % mod
    j = ((cont1 * valor1) + (cont2 * valor3) + (saida * valor2)) % mod
    k = ((cont1 * valor2) + (cont2 * valor1) + (saida * valor3)) % mod
    cont1 = i
    cont2 = j
    saida = k

print(saida)
     	  	     		  		 	 					