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())
MOD = 10**9+7
l -= 1
lrem = [0, 0, 0]
rrem = [0, 0, 0]
rem = [0, 0, 0]
if l%3 == 0:
    lrem[0] = l//3
    lrem[1] = l//3
    lrem[2] = l//3
elif l%3 == 1:
    lrem[0] = l//3
    lrem[1] = l//3 + 1
    lrem[2] = l//3
else: 
    lrem[0] = l//3
    lrem[1] = l//3 + 1
    lrem[2] = l//3 + 1
if r%3 == 0:
    rrem[0] = r//3
    rrem[1] = r//3
    rrem[2] = r//3
elif r%3 == 1:
    rrem[0] = r//3
    rrem[1] = r//3 + 1
    rrem[2] = r//3
else:
    rrem[0] = r//3
    rrem[1] = r//3 + 1
    rrem[2] = r//3 + 1
    
rem[0] = rrem[0]-lrem[0]
rem[1] = rrem[1]-lrem[1]
rem[2] = rrem[2]-lrem[2]

dp = [[0 for i in range(n)] for i in range(3)]
dp[0][0] = rem[0]
dp[1][0] = rem[1]
dp[2][0] = rem[2]

for i in range(1, n):
    dp[0][i] = (dp[0][i-1]*rem[0] + dp[1][i-1]*rem[2] + dp[2][i-1]*rem[1])%MOD
    dp[1][i] = (dp[1][i-1]*rem[0] + dp[2][i-1]*rem[2] + dp[0][i-1]*rem[1])%MOD
    dp[2][i] = (dp[2][i-1]*rem[0] + dp[1][i-1]*rem[1] + dp[0][i-1]*rem[2])%MOD
print(dp[0][n-1])
    