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
input = sys.stdin.readline
n,l,r = list(map(int,input().strip().split()))
mod = 1000000007
x=r-l+1
t=x//3

if(x%3==0):
    m0,m1,m2 = t,t,t
elif(x%3==1):
    m0,m1,m2 = t+1,t,t
elif(x%3==2):
    m0,m1,m2 = t+1,t+1,t

if(l%3==0):
    m0,m1,m2 = m0,m1,m2
elif(l%3==1):
    m1,m2,m0 = m0,m1,m2
elif(l%3==2):
    m2,m0,m1 = m0,m1,m2

dp = [ [0 for j in range(3)] for i in range(1)]
dp[0][0]=m0
dp[0][1]=m1
dp[0][2]=m2
for i in range(1,n):
    a=(dp[i-1][0]*m0%mod + dp[i-1][1]*m2%mod + dp[i-1][2]*m1%mod)%mod
    b=(dp[i-1][0]*m1%mod + dp[i-1][1]*m0%mod + dp[i-1][2]*m2%mod)%mod
    c=(dp[i-1][0]*m2%mod + dp[i-1][1]*m1%mod + dp[i-1][2]*m0%mod)%mod
    dp.append([a,b,c])

#print(dp)
print(dp[n-1][0])