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].
import math
mod = 10**9+7
n, L, R = map(int, input().strip().split())

ini = math.ceil(L/3)
end = math.floor(R/3)

n0 = end-ini+1

n1 = math.floor((R-1)/3)-math.ceil((L-1)/3)+1
n2 = math.floor((R-2)/3)-math.ceil((L-2)/3)+1

dp = [[0 for i in range(n)] for i in range(3)]

# print(n0, n1, n2)

dp[0][0] = n0
dp[1][0] = n1
dp[2][0] = n2

for i in range(1, n):
    dp[0][i] = (dp[1][i-1]*n2 + dp[2][i-1]*n1+dp[0][i-1]*n0)%mod
    dp[1][i] = (dp[0][i-1]*n1 + dp[2][i-1]*n2+dp[1][i-1]*n0)%mod
    dp[2][i] = (dp[0][i-1]*n2 + dp[1][i-1]*n1+dp[2][i-1]*n0)%mod

# for i in dp:
#     print(i)
print(dp[0][-1])


