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 itertools import combinations
from sys import stdin
import math
MOD=10**9+7

def listIn():
    return list((map(int,stdin.readline().strip().split())))

def stringIn():
    return([x for x in stdin.readline().split()])
    
def intIn():
    return (int(stdin.readline()))

n,l,r=listIn()
m1,m2=r%3,l%3
r-=m1
if m2!=0:
    l+=(3-m2)
t=(r-l)//3
mz,mo,mt=t+1,t,t
if m1==1:
    mo+=1
elif m1==2:
    mo+=1
    mt+=1
if m2==2:
    mt+=1
elif m2==1:
    mt+=1
    mo+=1

#print(mz,mo,mt)
dp=[]
for i in range(n+1):
    temp=[0]*3
    dp.append(temp)
    
dp[1][0],dp[1][1],dp[1][2]=mz,mo,mt
for i in range(2,n+1):
    dp[i][0]=(dp[i-1][0]*mz+dp[i-1][1]*mt+dp[i-1][2]*mo)%MOD
    dp[i][1]=(dp[i-1][0]*mo+dp[i-1][1]*mz+dp[i-1][2]*mt)%MOD
    dp[i][2]=(dp[i-1][0]*mt+dp[i-1][1]*mo+dp[i-1][2]*mz)%MOD
    
print(dp[n][0]%MOD)

