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].
MOD = 10 ** 9 + 7

n, l, r = [int(x) for x in input().strip().split()]
c = [0, 0, 0]

def f(x):
    y = x // 3
    if x % 3 == 0:
        return [y, y, y]
    elif x % 3 == 1:
        return [y, y+1, y]
    else:
        return [y, y+1, y+1]

c0 = f(l)
c1 = f(r)

c = [x - y for x,y in zip(c1, c0)]
c[l%3] += 1

r = c[:]

def mul_mod(x, y):
    return ((x % MOD) * (y % MOD))%MOD

for i in range(n-1):
    t = [0, 0, 0]
    t[0] = (mul_mod(r[0], c[0]) + mul_mod(r[1], c[2]) + mul_mod(r[2], c[1])) % MOD
    t[1] = (mul_mod(r[0], c[1]) + mul_mod(r[1], c[0]) + mul_mod(r[2], c[2])) % MOD
    t[2] = (mul_mod(r[0], c[2]) + mul_mod(r[1], c[1]) + mul_mod(r[2], c[0])) % MOD
    r = t[:]

print(r[0] % MOD)