1105_C. Ayoub and Lost Array
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].
```python
MOD = 10**9 + 7
n,l,r = map(int,input().split())

dp = [[ 0,0,0]for i in range(n+1)]

# In condition l <= x <= r , mod 0,1,2 number count(define a,b,c)
a = r//3     - (l - 1)//3
b = (r+1)//3 - (l - 1 + 1)//3
c = (r+2)//3 - (l - 1 + 2)//3
#example
#l = 2 , r = 10
#a = 10//3 - 1//3 = 3 - 0 = 3 (3,6,9)
#b = 11//3 - 2//3 = 3 - 0 = 3 (4,7,10)
#c = 12//3 - 3//3 = 4 - 1 = 3 (2,5,8)

dp[1][0] = a
dp[1][1] = b
dp[1][2] = c
for i in range(1,n):
    dp[i+1][0] = dp[i][0]*a%MOD + dp[i][1]*c%MOD + dp[i][2]*b%MOD
    dp[i+1][1] = dp[i][0]*b%MOD + dp[i][1]*a%MOD + dp[i][2]*c%MOD
    dp[i+1][2] = dp[i][0]*c%MOD + dp[i][1]*b%MOD + dp[i][2]*a%MOD

print(dp[n][0]%MOD)





```