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].
import math
M = 10**9+7
n,l,r = map(int,input().split())
e1 = math.ceil(l/3)
e2 = r//3
nz = e2-e1+1
e1 = math.ceil((l-1)/3)
e2 = (r-1)//3
no = e2-e1+1
e1 = math.ceil((l-2)/3)
e2 = (r-2)//3
nt = e2-e1+1
dp = []
for i in range(n):
    o = []
    for k in range(3):
        o.append(0)

    dp.append(o)

dp[0][0] = nz
dp[0][1] = no
dp[0][2] = nt
for i in range(1,n):
    dp[i][0] = ((dp[i-1][0]*nz)%M + (dp[i-1][1]*nt)%M + (dp[i-1][2]*no)%M)%M
    dp[i][1] = ((dp[i-1][0]*no)%M + (dp[i-1][1]*nz)%M + (dp[i-1][2]*nt)%M)%M
    dp[i][2] = ((dp[i-1][0]*nt)%M + (dp[i-1][1]*no)%M + (dp[i-1][2]*nz)%M)%M

print(dp[-1][0])

