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,a,b=[int(x) for x in input().split()]
c0=c1=c2=((b-a+1)//3)
x=c0*3
print
for item in range(a+x,b+1):
    if item%3==0:
        c0+=1
    elif item%3==1:
        c1+=1
    else:
        c2+=1

dp=[[0]*3 for i in range(n)]
dp[0][0]=c0
dp[0][1]=c1
dp[0][2]=c2
for i in range(1,n):
    dp[i][0]=(dp[i-1][0]*c0)%(10**9+7)+(dp[i-1][1]*c2)%(10**9+7)+(dp[i-1][2]*c1)%(10**9+7)
    dp[i][1]=(dp[i-1][0]*c1)%(10**9+7)+(dp[i-1][1]*c0)%(10**9+7)+(dp[i-1][2]*c2)%(10**9+7)
    dp[i][2]=(dp[i-1][0]*c2)%(10**9+7)+(dp[i-1][1]*c1)%(10**9+7)+(dp[i-1][2]*c0)%(10**9+7)
print((dp[n-1][0])%(10**9+7))
