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())
ost0 = r // 3 - (l - 1) // 3
ost1 = ost0
if l % 3 == 0:
    ost2 = ost0 - 1
if l % 3 == 1:
    ost2 = ost0
if l % 3 == 2:
    ost2 = ost0
if r % 3 == 2:
    ost2 += 1
if l % 3 == 0:
    ost1 = ost2
if l % 3 == 1:
    ost1 = ost2
if l % 3 == 2:
    ost1 = ost2 - 1
if r % 3 == 1:
    ost1 += 1
dp = []
for i in range(n):
    dp.append([0, 0, 0])
dp[0][0], dp[0][1], dp[0][2] = ost0, ost1, ost2
mmm = [ost0, ost1, ost2]
#dp[ost][n] = dp[0][n - 1] * mmm[ost] + dp[1][n - 1] * mmm[(ost - 1) % 3] + dp[2][n - 1] * mmm[(ost - 2) % 3]
for i in range(1, n):
    for j in range(3):
        dp[i][j] = (dp[i - 1][0] * mmm[j] + dp[i - 1][1] * mmm[(j - 1) % 3] + dp[i - 1][2] * mmm[(j - 2) % 3]) % (10 ** 9 + 7)
print(dp[-1][0] % (10 ** 9 + 7))