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 sys import stdin
from collections import Counter
n, l, r = map(int, input().split())
# dp = [[0] * (n + 1) for _ in range(3)]
c = [0] * 3
for i in range(l, min(r + 1, l + 3)):
    c[i % 3] = (r - i) // 3 + 1
MOD = 1000000007
dp0, dp1, dp2 = c[0], c[1], c[2]
for i in range(2, n + 1):
    dp0, dp1, dp2 = (dp0 * c[0] + dp1 * c[2] + dp2 * c[1]) % MOD, (
        dp0 * c[1] + dp1 * c[0] + dp2 * c[2]) % MOD, (
            dp0 * c[2] + dp1 * c[1] + dp2 * c[0]) % MOD

print(dp0)