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(' '))

m=1000000007

dp = [[0]*3 for _ in range(n+1)]

nums_0 = r//3-(l-1)//3
nums_1 = (r-1)//3 - (l-2)//3 
nums_2 = (r+1)//3 - (l)//3 

dp[1][0]=nums_0 %m
dp[1][1]=nums_1 %m
dp[1][2]=nums_2 %m

for j in range(2, n+1):
	dp[j][0] = (((dp[j-1][0] %m)*(nums_0 %m) %m) + ((dp[j-1][1] %m)*(nums_2 %m) %m) + ((dp[j-1][2] %m)*(nums_1 %m) %m) %m)
	dp[j][1] = (((dp[j-1][1] %m)*(nums_0 %m) %m) + ((dp[j-1][2] %m)*(nums_2 %m) %m) + ((dp[j-1][0] %m)*(nums_1 %m) %m) %m)
	dp[j][2] = (((dp[j-1][2] %m)*(nums_0 %m) %m) + ((dp[j-1][0] %m)*(nums_2 %m) %m) + ((dp[j-1][1] %m)*(nums_1 %m) %m) %m)


print(dp[n][0]%m)

