human
Write me a Python code to find the number of ways to restore the array given the following:
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].
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python code to find the number of ways to restore the array.

import math
n,l,r = map(int, input().split(' '))
zero = math.floor(r/3)-math.ceil(l/3)+1
one = math.floor((r-1)/3)-math.ceil((l-1)/3)+1
two = math.floor((r-2)/3)-math.ceil((l-2)/3)+1
mod=10**9+7

class Matrix:
    def __init__(self):
        self.m = [[0 for _ in range(3)] for j in range(3)]
    def __str__(self):
        print(self.m)
        return ""

def mul(a, b):
    ret = Matrix()
    for i in range(3):
        for j in range(3):
            ret.m[i][j] = 0
            for k in range(3):
                ret.m[i][j] += a.m[i][k] * b.m[k][j]
                ret.m[i][j] %= mod
    return ret

def pow(a, b):
    ret = Matrix()
    ret.m[0][0] = 1
    ret.m[1][1] = 1
    ret.m[2][2] = 1
    while b >= 1:
        if b % 2 == 1:
            ret = mul(ret, a)
        b >>= 1
        a = mul(a, a)
    return ret

a = Matrix()
a.m[0][0], a.m[1][1], a.m[2][2] = zero, zero, zero
a.m[0][2], a.m[1][0], a.m[2][1] = one, one, one
a.m[0][1], a.m[1][2], a.m[2][0] = two, two, two
ret = pow(a, n-1)
ans = ret.m[0][0] * zero + ret.m[0][1] * one + ret.m[0][2] * two
ans %= mod
print(ans)