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())
modz=r//3 - (l-1)//3
modo=(r+2)//3 - (l+1)//3
modt=(r+1)//3 - l//3
mod=1000000007
dp=[None]*(n+1)
for i in range(n+1):
    dp[i]=[0]*3
dp[1][0]=modz
dp[1][1]=modo
dp[1][2]=modt
for i in range(2,n+1):
        dp[i][0]=(dp[i-1][0]*modz + dp[i-1][1]*modt + dp[i-1][2]*modo)%mod
        dp[i][1]=(dp[i-1][0]*modo + dp[i-1][1]*modz + dp[i-1][2]*modt)%mod
        dp[i][2]=(dp[i-1][0]*modt + dp[i-1][1]*modo + dp[i-1][2]*modz)%mod
print(dp[n][0])
#print(dp)
#print(modz,modo,modt)