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].
[n, l, r] = [int(i) for i in input().split(" ")]

answer = 0
mod = []
if l == r:
    if (n * l) % 3 == 0:
        answer = 1
else:
    surplus = (r - l + 1) % 3
    mod = [int((r - l + 1 - surplus) / 3)] * 3
    if surplus == 1:
        mod[r % 3] += 1
    if surplus == 2:
        mod[r % 3] += 1
        mod[(r - 1) % 3] += 1
    dp = [([0] * (n + 1)) for i in range(3)]
    
    dp[0][0] = 1
    for i in range(1, n+1):    
        for j in range(3):
            dp[j][i] = (dp[j][i - 1] * mod[0] + dp[(j + 1)%3][i - 1] * mod[2] + dp[(j + 2)%3][i - 1] * mod[1]) % (1000000007)
    
    answer = dp[0][n]
print(answer)