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 sys
mod = 10**9+7

def LI(): return [int(x) for x in sys.stdin.readline().split()]

def main():
    n,l,r = LI()
    a = [1,0,0]
    t = [0,0,0]
    k = r // 3
    t = [k] * 3
    if r % 3 > 0:
        t[1] += 1
    if r % 3 > 1:
        t[2] += 1
    k = l // 3
    t = [c - k for c in t]
    if l % 3 == 0:
        t[0] += 1
    if l % 3 > 1:
        t[1] -= 1

    for _ in range(n):
        u = [0] * 3
        for i in range(3):
            k = a[i]
            for j in range(3):
                u[(i+j)%3] += k * t[j] % mod
        a = [c % mod for c in u]

    return a[0]



print(main())

