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(x) for x in input().split()]



diff_3 = (r - l)//3
ways = [diff_3,diff_3,diff_3]

for i in range(l,r-diff_3*3+1):
    # print(i)
    ways[i%3] += 1


# print(ways)
# print(sum(ways), r-l + 1)

curr = [ways[0],ways[1],ways[2]]
# print(curr)
for i in range(n-1):
    new = [0,0,0]
    for j in range(0,3):
        for k in range(0,3):
            new[(j+k)%3] += (ways[j]*curr[k])
    for j in range(0,3):
        new[j] %= (10**9+7)
    curr = new
    # print(curr)

print(curr[0])