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].
from sys import stdin
n,l,r=map(int,stdin.readline().strip().split())
dp=[[0 for i in range(3)]for j in range(n)]
x=l//3
if l%3!=0:
    x+=1
y=3*x
z=0

for i in range(l,y+1):
    z+=1
    if i%3==0:
        dp[0][0]+=1
    if i%3==1:
        dp[0][1]+=1
    if i%3==2:
        dp[0][2]+=1
x=r//3
y=3*x

for i in range(y+1,r+1):
    z+=1
    if i%3==0:
        dp[0][0]+=1
    if i%3==1:
        dp[0][1]+=1
    if i%3==2:
        dp[0][2]+=1
x=(r-l+1-z)//3
mod=10**9+7
dp[0][0]+=x
dp[0][1]+=x
dp[0][2]+=x
for i in range(1,n):
    dp[i][0]=((dp[0][0]*dp[i-1][0])%mod+(dp[0][1]*dp[i-1][2])%mod+(dp[0][2]*dp[i-1][1])%mod)%mod
    dp[i][1]=((dp[0][0]*dp[i-1][1])%mod+(dp[0][1]*dp[i-1][0])%mod+(dp[0][2]*dp[i-1][2])%mod)%mod
    dp[i][2]=((dp[0][1]*dp[i-1][1])%mod+(dp[0][0]*dp[i-1][2])%mod+(dp[0][2]*dp[i-1][0])%mod)%mod
print(dp[-1][0])
    
    
