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].
#  =========     /\       /|    |====/|
#      |        /  \       |    |   / |
#      |       /____\      |    |  /  |
#      |      /      \     |    | /   |
#  ========= /        \  =====  |/====|  
#  code

if __name__ == "__main__":
    n,l,r = map(int,input().split())
    dp = [[0,0,0] for i in range(n)]
    MOD = int(1e9 + 7)

    t0 = r//3 - (l - 1)//3
    t1 = (r - 1)//3 - (l + 1)//3 + 1
    t2 = r - l - t0 - t1 + 1

    dp[0][0] = t0%MOD
    dp[0][1] = t1%MOD
    dp[0][2] = t2%MOD

    i = 1
    while i < n:
        dp[i][0] = (dp[i-1][0] * t0 + dp[i-1][1] * t2 + dp[i-1][2] * t1)%MOD
        dp[i][1] = (dp[i-1][0] * t1 + dp[i-1][1] * t0 + dp[i-1][2] * t2)%MOD
        dp[i][2] = (dp[i-1][0] * t2 + dp[i-1][1] * t1 + dp[i-1][2] * t0)%MOD
        i += 1
    print(dp[n-1][0]%MOD)


