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].
# -*- coding: utf-8 -*-
# @Time    : 2019/1/20 20:36
# @Author  : LunaFire
# @Email   : gilgemesh2012@gmail.com
# @File    : C. Ayoub and Lost Array.py


def main():
    n, l, r = map(int, input().split())
    m = 10 ** 9 + 7

    sub = r - l + 1
    num = [sub // 3] * 3
    if sub % 3 == 1:
        num[l % 3] += 1
    elif sub % 3 == 2:
        num[l % 3] += 1
        num[(l + 1) % 3] += 1

    dp = [1, 0, 0]
    for _ in range(n):
        temp = [0] * 3
        temp[0] = dp[0] * num[0] + dp[1] * num[2] + dp[2] * num[1]
        temp[1] = dp[0] * num[1] + dp[1] * num[0] + dp[2] * num[2]
        temp[2] = dp[0] * num[2] + dp[1] * num[1] + dp[2] * num[0]
        for i in range(3):
            dp[i] = temp[i] % m

    print(dp[0])


if __name__ == '__main__':
    main()
