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].
mod2 = 10**9 + 7
mod = mod2 * 3
n,l,r = map(int,input().split())
l2 = l // 3 * 3
r2 = (r // 3 + 1) * 3 - 1
num0 = r2 - l2  + 1
num0 //= 3
nums = [num0,num0,num0]
for i in range(l2,l):
    nums[i % 3]  -= 1
for i in range(r+1,r2+1):
    nums[i % 3]  -= 1
ar = [[0,0,0] for i in range(n)]
ar[0] = nums
for i in range(n-1):
    for j in range(3):
        for z in range(3):
            ar[i + 1][(j+z) % 3] +=  ar[i][j] * nums[z]
            ar[i + 1][(j+z) % 3] %= mod2
print(ar[-1][0])
