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].
from math import ceil
n, l, r = list(map(int, input().split()))
m0, m1, m2 = 0, 0, 0
mod=1000000007
m0=r//3-(l-1)//3
m1=(r-1)//3-(l-2)//3
m2=(r-2)//3-(l-3)//3
# print(m0,m1,m2)
dp = [[0 for _ in range(3)] for _ in range(n)]
dp[0][0] = m0
dp[0][1] = m1
dp[0][2] = m2
for i in range(1, n):
    dp[i][0] =((dp[i - 1][0] * m0)%mod + (dp[i - 1][1] * m2)%mod+ (dp[i - 1][2] * m1)%mod)%mod
    dp[i][1] = ((dp[i - 1][0] * m1)%mod + (dp[i - 1][1] * m0)%mod + (dp[i - 1][2] * m2)%mod)%mod
    dp[i][2] = ((dp[i - 1][0] * m2)%mod + (dp[i - 1][1] * m1)%mod + (dp[i - 1][2] * m0)%mod)%mod
print(dp[n-1][0]%mod)
