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

def main():
    n, l, r = (int(a) for a in input().split())
    mod = 10**9 + 7
    nums = [(math.floor(r/3) - math.ceil(l/3) + 1) % mod,
            (math.floor((r-1)/3) - math.ceil((l-1)/3) + 1) % mod,
            (math.floor((r-2)/3) - math.ceil((l-2)/3) + 1) % mod]
    rem = nums[:]
    for i in range(n-1):
        newRem = []
        # Remainder is 0
        # 0 + 0 / 1 + 2 / 2 + 1
        newRem.append(
            ((nums[0] * rem[0]) % mod +
             (nums[1] * rem[2]) % mod +
             (nums[2] * rem[1]) % mod) % mod
        )
        # Remainder is 1
        # 0 + 1 / 1 + 0 / 2 + 2
        newRem.append(
            ((nums[0] * rem[1]) % mod +
             (nums[1] * rem[0]) % mod +
             (nums[2] * rem[2]) % mod) % mod
        )
        # Remainder is 2
        # 0 + 2 / 1 + 1 / 2 + 0
        newRem.append(
            ((nums[0] * rem[2]) % mod +
             (nums[1] * rem[1]) % mod +
             (nums[2] * rem[0]) % mod) % mod
        )
        rem = newRem[:]
    print(rem[0])

main()
    
