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].
from collections import *
from math import *

n, l, r = map(int, input().split())
Mod = int(1e9+7)

dp = [[0 for i in range(3)] for j in range(n+1)]
m0 = floor(r/3) - ceil(l/3) + 1
m1 = floor((r-1)/3) - ceil((l-1)/3) + 1
m2 = floor((r-2)/3) - ceil((l-2)/3) + 1
dp[0][0] = 1

for i in range(1, n+1):
    dp[i][0] = dp[i-1][0]*m0 + dp[i-1][1]*m2 + dp[i-1][2]*m1
    dp[i][1] = dp[i-1][1]*m0 + dp[i-1][2]*m2 + dp[i-1][0]*m1
    dp[i][2] = dp[i-1][2]*m0 + dp[i-1][0]*m2 + dp[i-1][1]*m1
    for j in range(3):
        dp[i][j] %= Mod

print(dp[n][0])
# print(m0, m1, m2)