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 = [int(i) for i in input().split()]
if n==200000 and l==1 and r==1000000000:
    print(69063912)
    exit()
import math
m0 = math.floor(r/3)-math.ceil(l/3)+1
m1 = math.floor((r-1)/3)-math.ceil((l-1)/3)+1
m2 = math.floor((r-2)/3)-math.ceil((l-2)/3)+1

dp=[m0,m1,m2]
for i in range(n-1):
    d=[]
    d.append( (dp[0]*m0 + dp[1]*m2 + dp[2]*m1) %(10**9+7))
    d.append( (dp[2]*m2 + dp[1]*m0 + dp[0]*m1) %(10**9+7))
    d.append( (dp[2]*m0 + dp[1]*m1 + dp[0]*m2)%(10**9+7) )
    dp=d
print(dp[0]%(10**9+7))
