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, r, l=(int(i) for i in input().split())
r0=r
if r0%3==1: r0+=2
elif r0%3==2: r0+=1
l0=l
if l0%3==1: l0-=1
elif l0%3==2: l0-=2
d0=(l0-r0)//3+1
d1=d0-1
d2=d0-1
if r0-r==2:
    d1+=1
    d2+=1
elif r0-r==1:
    d2+=1
if l-l0==2:
    d1+=1
    d2+=1
elif l-l0==1:
    d1+=1
d0=[d0]
d1=[d1]
d2=[d2]
#print(d0[0], d1[0], d2[0])
for i in range(1, n):
    d0.append((d0[i-1]*d0[0]+d1[i-1]*d2[0]+d2[i-1]*d1[0])%1000000007)
    d1.append((d0[i-1]*d1[0]+d1[i-1]*d0[0]+d2[i-1]*d2[0])%1000000007)
    d2.append((d0[i-1]*d2[0]+d1[i-1]*d1[0]+d2[i-1]*d0[0])%1000000007)
print(d0[n-1])
