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=[int(x) for x in input().split()]
m=1000000007
if r%3==0:
    count1=r//3
    count2=r//3
    count3=r//3
elif r%3==1:
    count1=r//3 +1
    count2=r//3
    count3=r//3
else:
    count1=r//3 +1
    count2=r//3 +1
    count3=r//3
rcount1=count1
rcount2=count2
rcount0=count3

l-=1

if l%3==0:
    count1=l//3
    count2=l//3
    count3=l//3
elif l%3==1:
    count1=l//3 +1
    count2=l//3
    count3=l//3
else:
    count1=l//3 +1
    count2=l//3 +1
    count3=l//3
lcount1=count1
lcount2=count2
lcount0=count3

finalcount1=rcount1-lcount1
finalcount2=rcount2-lcount2
finalcount0=rcount0-lcount0

dp0=[0]*200001
dp1=[0]*200001
dp2=[0]*200001

dp0[0]=finalcount0
dp1[0]=finalcount1
dp2[0]=finalcount2
# print(finalcount0,finalcount1,finalcount2)
for i in range(1,n):
    dp0[i]=(dp0[i-1]*finalcount0)%m
    dp0[i]=(dp0[i]+(dp1[i-1]*finalcount2)%m)%m
    dp0[i]=(dp0[i]+(dp2[i-1]*finalcount1)%m)%m

    dp1[i]=(dp1[i-1]*finalcount0)%m
    dp1[i]=(dp1[i]+(dp0[i-1]*finalcount1)%m)%m
    dp1[i]=(dp1[i]+(dp2[i-1]*finalcount2)%m)%m

    dp2[i]=(dp0[i-1]*finalcount2)%m
    dp2[i]=(dp2[i]+(dp1[i-1]*finalcount1)%m)%m
    dp2[i]=(dp2[i]+(dp2[i-1]*finalcount0)%m)%m
ans=dp0[n-1]
print(ans)