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())

mod = (10 ** 9) + 7

new_l = 0
new_r = 0

if l % 3 == 0:
    new_l = l
elif l % 3 == 1:
    new_l = l - 1
else:
    new_l = l - 2

if r % 3 == 2:
    new_r = r
elif r % 3 == 1:
    new_r = r + 1
else:
    new_r = r + 2

def count(mod):
    f_l = lambda x: sum(1 if i % 3 == x else 0 for i in range(new_l, l))
    f_r = lambda x: sum(1 if i % 3 == x else 0 for i in range(r + 1, new_r + 1))
    return (new_r - new_l + 1) // 3 - f_r(mod) - f_l(mod)

init = {
    0: count(0),
    1: count(1),
    2: count(2),
}

a = {
    0: init[0],
    1: init[1],
    2: init[2],
}


for i in range(1, n):
    a0 = (a[0] * init[0] + a[2] * init[1] + a[1] * init[2]) % mod
    a1 = (a[1] * init[0] + a[0] * init[1] + a[2] * init[2]) % mod
    a2 = (a[2] * init[0] + a[1] * init[1] + a[0] * init[2]) % mod
    a[0] = a0
    a[1] = a1
    a[2] = a2

print(a[0])
