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 = map(int, input().split())
mod = 1000000007

ans = [[0] * n, [0] * n, [0] * n]
ans[0][0] = r // 3 - (l - 1) // 3
ans[1][0] = (r + 2) // 3 - (l + 1) // 3
ans[2][0] = (r + 1) // 3 - l // 3

if n == 1:
    print(ans[0][0])
    exit(0)

for i in range(1, n - 1):
    ans[0][i] = (ans[0][i - 1] * ans[0][0] + ans[1][i - 1] * ans[2][0] + ans[2][i - 1] * ans[1][0]) % mod
    ans[1][i] = (ans[0][i - 1] * ans[1][0] + ans[1][i - 1] * ans[0][0] + ans[2][i - 1] * ans[2][0]) % mod
    ans[2][i] = (ans[0][i - 1] * ans[2][0] + ans[1][i - 1] * ans[1][0] + ans[2][i - 1] * ans[0][0]) % mod

ans[0][n - 1] = (ans[0][n - 2] * ans[0][0]) % mod
ans[1][n - 1] = (ans[1][n - 2] * ans[2][0]) % mod
ans[2][n - 1] = (ans[2][n - 2] * ans[1][0]) % mod

print((ans[0][n - 1] + ans[1][n - 1] + ans[2][n - 1]) % mod)