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())
z1=[0]*3
s=l%3
s1=r%3
if s==s1:
    z1[s]+=1
    for i in range(3):
        z1[i]+=(r-l)//3
elif s-1==s1 or s==0 and s1==2:
    for i in range(3):
        z1[i]+=(r-l+1)//3
elif (s+1)%3==s1:
    for i in range(3):
        z1[i]+=(r-l+2)//3
    z1[(s1+1)%3]-=1
dp=[[0 for j in range(3)]for i in range(n)]
x=z1[0]
y=z1[1]
z=z1[2]
dp[0][0]=x
dp[0][1]=y
dp[0][2]=z
for i in range(1,n):
    dp[i][0]=(dp[i-1][0]*x+dp[i-1][1]*z+dp[i-1][2]*y)%1000000007
    dp[i][1]=(dp[i-1][0]*y+dp[i-1][1]*x+dp[i-1][2]*z)%1000000007
    dp[i][2]=(dp[i-1][0]*z+dp[i-1][1]*y+dp[i-1][2]*x)%1000000007
print(dp[n-1][0])