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())
a=r//3 - (l-1)//3
l1=l
r1=r
mod=10**9 + 7
if r1%3!=2:
    r1-= ((r1%3) + 1)
if l1%3!=2:
    l1+= (-(l1%3) + 2)

b=(r1 - l1)//3 + 1

l1=l
r1=r
if r1%3!=1:
    r1-= (2-(r1%3) )
if l1%3==2:
    l1+=2
elif l1%3==0:
    l1+=1
c=(r1-l1)//3 + 1
a%=mod
b%=mod
c%=mod


dp=[[a,c,b] for i in range(n+1)]

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

print(dp[n][0])

    
    
    