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 = [int(i) for i in input().split()]

#dp = [[0 for i in range(3)] for j in range(3)]

number_of_zeros = 0
number_of_ones = 0
number_of_twos = 0

if l%3 == 0:
    if r%3 == 0:
        number_of_zeros = (r-l+1)//3 + 1
        number_of_ones = (r-l+1)//3
        number_of_twos = (r-l+1)//3
    if r%3 == 1:
        number_of_zeros = (r-l+1)//3 + 1
        number_of_ones = (r-l+1)//3 + 1
        number_of_twos = (r-l+1)//3
    if r%3 == 2:
        number_of_zeros = (r-l+1)//3
        number_of_ones = (r-l+1)//3
        number_of_twos = (r-l+1)//3
elif l%3 == 1:
    if r%3 == 0:
        number_of_zeros = (r-l+1)//3
        number_of_ones = (r-l+1)//3
        number_of_twos = (r-l+1)//3
    if r%3 == 1:
        number_of_zeros = (r-l+1)//3
        number_of_ones = (r-l+1)//3 + 1
        number_of_twos = (r-l+1)//3
    if r%3 == 2:
        number_of_zeros = (r-l+1)//3
        number_of_ones = (r-l+1)//3 + 1
        number_of_twos = (r-l+1)//3 + 1
elif l%3 == 2:
    if r%3 == 0:
        number_of_zeros = (r-l+1)//3 + 1
        number_of_ones = (r-l+1)//3
        number_of_twos = (r-l+1)//3 + 1
    if r%3 == 1:
        number_of_zeros = (r-l+1)//3
        number_of_ones = (r-l+1)//3
        number_of_twos = (r-l+1)//3
    elif r%3 == 2:
        number_of_zeros = (r-l+1)//3 
        number_of_ones = (r-l+1)//3
        number_of_twos = (r-l+1)//3 + 1

dp = [[-1 for i in range(3)] for j in range(n+1)]

MOD = 10**9 + 7
dp[1][0] = number_of_zeros%MOD
dp[1][1] = number_of_ones%MOD
dp[1][2] = number_of_twos%MOD

for i in range(2,n+1):
    dp[i][0] = (((dp[i-1][0]%MOD)*(number_of_zeros%MOD))%MOD + ((dp[i-1][1]%MOD)*(number_of_twos%MOD))%MOD + ((dp[i-1][2]%MOD)*(number_of_ones%MOD))%MOD)%MOD
    dp[i][1] = (((dp[i-1][0]%MOD)*(number_of_ones%MOD))%MOD + ((dp[i-1][1]%MOD)*(number_of_zeros%MOD))%MOD + ((dp[i-1][2]%MOD)*(number_of_twos%MOD))%MOD)%MOD
    dp[i][2] = (((dp[i-1][0]%MOD)*(number_of_twos%MOD))%MOD + ((dp[i-1][1]%MOD)*(number_of_ones%MOD))%MOD + ((dp[i-1][2]%MOD)*(number_of_zeros)%MOD)%MOD)%MOD

if dp[n][0]!=-1:
    print(dp[n][0]%MOD)
else:
    print(0)
