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].
from collections import defaultdict
#n = int(raw_input())
n,l,r = [int(x) for x in raw_input().split()]

if l == r:
    print('1' if n % 3 == 0 or l % 3 == 0 else '0')
    exit()
mod = 10**9 + 7
lm = l % 3
rn = r
num = (r - l) // 3
d = {0:num,1:num,2:num}

while rn % 3 != lm:
    d[rn % 3] += 1
    rn -= 1
d[rn % 3] += 1
#print(d)

vars = {0:1,1:0,2:0}
for i in range(n):
    newvars = {0:0,1:0,2:0}
    newvars[0] = (vars[0] * d[0] + vars[1] * d[2] + vars[2] * d[1]) % mod
    newvars[1] = (vars[1] * d[0] + vars[0] * d[1] + vars[2] * d[2]) % mod
    newvars[2] = (vars[1] * d[1] + vars[0] * d[2] + vars[2] * d[0]) % mod
    vars = newvars
    #print(vars[0])
print(vars[0])
