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].
from math import *
n, l, r = map(int, input().split())
a = floor(r / 3) - ceil(l / 3) + 1
b = floor((r - 1) / 3) - ceil((l - 1) / 3) + 1
c = floor((r - 2) / 3) - ceil((l - 2) / 3) + 1
m = 1000000007
dp = [ [0] * (n + 1) for i in range(3)]
dp[0][1] = a
dp[1][1] = b
dp[2][1] = c
for l in range(2, n + 1):
    dp[0][l] = (dp[0][l - 1] * a + dp[1][l - 1] * c + dp[2][l - 1] * b) % m
    dp[1][l] = (dp[0][l - 1] * b + dp[1][l - 1] * a + dp[2][l - 1] * c) % m
    dp[2][l] = (dp[0][l - 1] * c + dp[1][l - 1] * b + dp[2][l - 1] * a) % m;
    
print(dp[0][n] % m)