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,stdout
M=10**9+7
for _ in range(1):#int(stdin.readline())):
    # n=int(stdin.readline())
    n,l,r=list(map(int,stdin.readline().split()))
    a0=a1=a2=l0=l1=l2=n0=n1=n2=0
    for i in range(l,min(l+3,r+1)):
        if i%3==0:a0=i
        elif i%3==1:a1=i
        else:a2=i
    for i in range(r,max(l-1,r-3),-1):
        if i%3==0:l0=i
        elif i%3==1:l1=i
        else:l2=i
    if a0!=0:n0=(l0-a0)//3+1
    if a1!=0:n1=(l1-a1)//3+1
    if a2!=0:n2=(l2-a2)//3+1
    dp0,dp1,dp2=n0,n1,n2
    # print(n0,n1,n2)
    for i in range(2,n+1):
        dp0,dp1,dp2=(dp0*n0+dp1*n2+dp2*n1)%M,(dp0*n1+dp1*n0+dp2*n2)%M,(dp0*n2+dp2*n0+n1*dp1)%M
    print(dp0%M)