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].
mod = 1000000007
n,l,r = map(int,input().split())
arr=[0]*3
while l%3!=0:
    arr[l%3]+=1
    l+=1
while r%3!=0:
    arr[r%3]+=1
    r-=1
aa=(r-l)//3  
arr[0]+=(aa+1)
arr[1]+=aa
arr[2]+=aa
dp=[[0,0,0] for i in range(n)]
dp[0][0]=arr[0];dp[0][1]=arr[1];dp[0][2]=arr[2]
for i in range(1,n):
    dp[i][0] = ((dp[i-1][0]*arr[0])%mod + (dp[i-1][1]*arr[2])%mod + (dp[i-1][2]*arr[1])%mod)%mod
    dp[i][1] = ((dp[i-1][0]*arr[1])%mod + (dp[i-1][1]*arr[0])%mod + (dp[i-1][2]*arr[2])%mod)%mod 
    dp[i][2] = ((dp[i-1][0]*arr[2])%mod + (dp[i-1][1]*arr[1])%mod + (dp[i-1][2]*arr[0])%mod)%mod            
print(dp[n-1][0]%mod)
