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].
#! usr/bin/env python
# -*- coding: utf-8 -*-


MOD = 10 ** 9 + 7
MAX_N = 2 * 10 ** 5 + 5
dp = [[0] * 3 for i in range(MAX_N)]


def main():
    n, l, r = map(int, input().split())
    k0 = r // 3 - (l-1) // 3
    k1 = (r//3+(r % 3 >= 1)) - ((l-1)//3+((l-1) % 3 >= 1))
    k2 = (r//3+(r % 3 >= 2)) - ((l-1)//3+((l-1) % 3 >= 2))

    dp[0][0], dp[0][1], dp[0][2] = k0, k1, k2
    for i in range(1, n):
        dp[i][0] = dp[i-1][0] * k0 + dp[i-1][1] * k2 + dp[i-1][2] * k1
        dp[i][1] = dp[i-1][0] * k1 + dp[i-1][1] * k0 + dp[i-1][2] * k2
        dp[i][2] = dp[i-1][0] * k2 + dp[i-1][1] * k1 + dp[i-1][2] * k0
        for j in range(3):
            dp[i][j] %= MOD

    print(dp[n-1][0])


if __name__ == '__main__':
    main()
