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 = map(int,input().split())

D = [0,0,0]

for i in range(l,l+(-l)%3):
    D[i%3] += 1
l = l + (-l)%3
for i in range(r-r%3,r+1):
    D[i%3] += 1
r = r - r%3 - 1

from math import factorial as F
def C(n,k):
    return F(n)//(F(n-k)*F(k))

D[0] += (r-l+1)//3
D[1] += (r-l+1)//3
D[2] += (r-l+1)//3

DP = {(-1,0): 1, (-1,1): 0, (-1,2): 0}
for i in range(n):
    for m in range(3):
        DP[i,m] = (DP[i-1,m]*D[0] +
                   DP[i-1,(m-1)%3]*D[1] + 
                   DP[i-1,(m-2)%3]*D[2]) % mod

print(DP[n-1,0])
