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].
R = lambda: map(int, input().split())
n, l, r = R()
lb, rb = [-1] * 3, [-1] * 3
for i in range(3):
    if l + i <= r:
        lb[(l + i) % 3] = l + i
    if r - i >= l:
        rb[(r - i) % 3] = r - i
bs = [0] * 3
for i in range(3):
    if lb[i] > 0 and rb[i] > 0:
        bs[i] = (rb[i] - lb[i]) // 3 + 1
    elif lb[i] > 0 or rb[i] > 0:
        bs[i] = 1
cnt = bs[:]
for i in range(n - 1):
    tmp = [0] * 3
    for j in range(3):
        for k in range(3):
            tmp[(j + k) % 3] += cnt[j] * bs[k]
    cnt = [x % (10**9 + 7) for x in tmp]
print(cnt[0])