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].
M = 10**9 + 7


def types(n):
    if n % 3 == 0:
        return n // 3, n // 3, n // 3
    elif n % 3 == 1:
        return n // 3, n // 3 + 1, n // 3
    return n // 3, n // 3 + 1, n // 3 + 1


n, l, r = map(int, input().split())

nr0, nr1, nr2 = types(r)
nl0, nl1, nl2 = types(l - 1)
n0, n1, n2 = nr0 - nl0, nr1 - nl1, nr2 - nl2

c0, c1, c2 = 1, 0, 0
for _ in range(n):
    c0n = c0 * n0 + c1 * n2 + c2 * n1
    c1n = c0 * n1 + c1 * n0 + c2 * n2
    c2n = c0 * n2 + c1 * n1 + c2 * n0
    c0, c1, c2 = c0n % M, c1n % M, c2n % M

print(c0)
