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, raw_input().split())
m = 10 ** 9 + 7

a = [0, 0, 0]

a[l % 3] = (r - l + 1) / 3
a[(l + 1) % 3] = (r - l + 1) / 3
a[(l + 2) % 3] = (r - l + 1) / 3

k = (r - l + 1) % 3
if k == 1: a[l % 3] += 1
if k == 2:
    a[l % 3] += 1
    a[(l + 1) % 3] += 1

dp = [a[0], a[1], a[2]]
for i in range(1, n):
    da = (dp[0] * a[0] + dp[1] * a[2] + dp[2] * a[1]) % m
    db = (dp[0] * a[1] + dp[1] * a[0] + dp[2] * a[2]) % m
    dc = (dp[0] * a[2] + dp[1] * a[1] + dp[2] * a[0]) % m
    dp = [da, db, dc]

print dp[0] % m
