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].
n,l,r=map(int,input().strip().split(' ')) 
a=[0,0,0] 
a[0]=a[1]=a[2]=(r-l+1)//3
if (r-l+1)%3==2:
    a[1]+=1 
    a[2]+=1 
elif (r-l+1)%3==1:
    a[1]+=1 
if l%3==2:
    a=[a[2],a[0],a[1]]
elif l%3==0:
    a=[a[1],a[2],a[0]] 
m=1000000007
now0=[0]*(n+1) 
now1=[0]*(n+1)
now2=[0]*(n+1)
now2[1]=a[2]
now0[1]=a[0]
now1[1]=a[1]
for i in range(2,n+1):
    now0[i]=(now0[i-1]*a[0]+now1[i-1]*a[2]+now2[i-1]*a[1])%m
    
    now1[i]=(now0[i-1]*a[1]+now1[i-1]*a[0]+now2[i-1]*a[2])%m
    now2[i]=(now0[i-1]*a[2]+now1[i-1]*a[1]+now2[i-1]*a[0])%m 
print(now0[n]%m)