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 sys
import math
import collections
import heapq
import decimal
input=sys.stdin.readline
n,l,r=(int(i) for i in input().split())
mod=1000000007
c1=r//3-math.ceil(l/3)+1
c2=(r-1)//3-math.ceil((l-1)/3)+1
c3=(r-2)//3-math.ceil((l-2)/3)+1
dp=[[0,0,0] for i in range(n+1)]
dp[1][0]=c1
dp[1][1]=c2
dp[1][2]=c3
for i in range(2,n+1):
    dp[i][0]=((dp[i-1][0]*c1)%mod+(dp[i-1][1]*c3)%mod+(dp[i-1][2]*c2)%mod)%mod
    dp[i][1]=((dp[i-1][0]*c2)%mod+(dp[i-1][1]*c1)%mod+(dp[i-1][2]*c3)%mod)%mod
    dp[i][2]=((dp[i-1][0]*c3)%mod+(dp[i-1][1]*c2)%mod+(dp[i-1][2]*c1)%mod)%mod
print(dp[n][0])