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().split())
c0 = (r//3*3-l//3*3)//3+1
c1 = (r//3*3-l//3*3)//3
c2 = (r//3*3-l//3*3)//3
if(l%3==1):
    c0-=1
elif(l%3==2):
    c0-=1
    c1-=1
if(r%3==1):
    c1+=1
elif(r%3==2):
    c2+=1
    c1+=1
cnt0 = [c0]
cnt1 = [c1]
cnt2 = [c2]
for i in range(n):
    cnt0.append((cnt0[i]*c0+cnt1[i]*c2+cnt2[i]*c1)%1000000007)
    cnt1.append((cnt0[i]*c1+cnt1[i]*c0+cnt2[i]*c2)%1000000007)
    cnt2.append((cnt0[i]*c2+cnt1[i]*c1+cnt2[i]*c0)%1000000007)
print(cnt0[-2]%1000000007)
