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].
import sys


def main():
    MOD = 1000 * 1000 * 1000 + 7
    n, l, r = [int(x) for x in sys.stdin.readline().strip().split()]

    dp = [[0] * 3 for i in range(n + 1)]

    rnum = r // 3
    lnum = l // 3
    if lnum + 1 < rnum:
        dp[1][0] = dp[1][1] = dp[1][2] = rnum - lnum - 1
        for ix in range(l, 3 * (lnum + 1)):
            dp[1][ix % 3] += 1
        for ix in range(3 * rnum, r + 1):
            dp[1][ix % 3] += 1
    else:
        for ix in range(l, r + 1):
            dp[1][ix % 3] += 1

    for sz in range(2, n + 1):
        for rem in range(0, 3):
            dp[sz][rem] = (
                (dp[sz - 1][rem] * dp[1][0]) % MOD
                + (dp[sz - 1][(rem - 1 + 3) % 3] * dp[1][1]) % MOD
                + (dp[sz - 1][(rem - 2 + 3) % 3] * dp[1][2]) % MOD
            ) % MOD
    print(dp[n][0])


if __name__ == '__main__':
    main()

