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].
import math
n,l,r=map(int,input().split(' '))
zero=math.floor(r/3)-math.ceil(l/3)+1
one=math.floor((r-1)/3)-math.ceil((l-1)/3)+1
two=math.floor((r-2)/3)-math.ceil((l-2)/3)+1
a=[zero,one,two]
mod=10**9+7
dp=[[1 for i in range(n)] for i in range(3)]
for i in range(n):
    for j in range(3):
        if i==0:
            dp[j][i]=a[j]
        else:
            if j==0:
                dp[j][i]=((dp[j][i-1]*a[0])+(dp[j+1][i-1]*a[2])+(dp[j+2][i-1]*a[1]))%mod
            elif j==1:
                dp[j][i]=((dp[j-1][i-1]*a[1])+(dp[j][i-1]*a[0])+(dp[j+1][i-1]*a[2]))%mod
            else:
                dp[j][i]=((dp[j-2][i-1]*a[2])+(dp[j-1][i-1]*a[1])+(dp[j][i-1]*a[0]))%mod
print(dp[0][n-1])
