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())
total = r-l+1
rf, re = l%3, r%3
if rf==0: rf = 3
l, r = l+(3-rf), r-re
if rf==3: rf = 0

mod = int(1e9+7)
zeros = ((r-l)//3 + 1)%mod
ones = ((rf==1)*1 + (r-l)//3 + (re>=1)*1)%mod
twos = ((rf>=1)*1 + (r-l)//3 + (re==2)*1)%mod

#dp[i][j], size of array whose sum modulo 3 is j
dp = [ [None]*3 for _ in range(n+1) ]

dp[1][0], dp[1][1], dp[1][2] = zeros, ones, twos

for i in range(2,n+1):
    dp[i][0] = (((dp[i-1][0]%mod)*zeros)%mod + ((dp[i-1][1]%mod)*twos)%mod + ((dp[i-1][2]%mod)*ones)%mod)%mod
    dp[i][1] = (((dp[i-1][0]%mod)*ones)%mod + ((dp[i-1][1]%mod)*zeros)%mod + ((dp[i-1][2]%mod)*twos)%mod)%mod
    dp[i][2] = (((dp[i-1][0]%mod)*twos)%mod + ((dp[i-1][1]%mod)*ones)%mod + ((dp[i-1][2]%mod)*zeros)%mod)%mod
print(dp[n][0])