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 = map(int, input().split())
a0 = [0 for i in range(n)]
a1 = [0 for i in range(n)]
a2 = [0 for i in range(n)]
k = r - l + 1
a0[0] = k // 3
a1[0] = k // 3
a2[0] = k // 3
if k % 3 == 1:
    if l % 3 == 1:
        a1[0] += 1
    elif l % 3 == 2:
        a2[0] += 1
    else:
        a0[0] += 1
elif k % 3 == 2:
    if l % 3 == 1:
        a1[0] += 1
        a2[0] += 1
    elif l % 3 == 2:
        a2[0] += 1
        a0[0] += 1
    else:
        a0[0] += 1
        a1[0] += 1
p0 = a0[0]
p1 = a1[0]
p2 = a2[0]
for i in range(1, n):
    a0[i] = (a0[i - 1] * p0 + a1[i - 1] * p2 + a2[i - 1] * p1) % (10**9 + 7)
    a1[i] = (a0[i - 1] * p1 + a1[i - 1] * p0 + a2[i - 1] * p2) % (10**9 + 7)
    a2[i] = (a0[i - 1] * p2 + a1[i - 1] * p1 + a2[i - 1] * p0) % (10**9 + 7)
print(a0[n - 1])