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 sys
# import heapq
from math import ceil,floor



RI = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
mod = 10**9+7
# for _ in range(int(ri())):
n,l,r = RI()
dic = {0:0,1:0,2:0}
n0 = [0]*n
n1 = [0]*n
n2 = [0]*n
dic[0] = floor(r/3)-ceil(l/3)+1
dic[1] = floor((r-1)/3)-ceil((l-1)/3)+1
dic[2] = floor((r-2)/3)-ceil((l-2)/3)+1
# for i in range(l,r+1):
#     dic[i%3]+=1
n0[0] = dic[0]
n1[0] = dic[1]
n2[0] = dic[2]
for i in range(1,n):
    n0[i] = ((n1[i-1]*dic[2])%mod + (n2[i-1]*dic[1])%mod + (n0[i-1]*dic[0])%mod)%mod
    n1[i] = ((n1[i-1]*dic[0])%mod + (n2[i-1]*dic[2])%mod + (n0[i-1]*dic[1])%mod)%mod
    n2[i] = ((n1[i-1]*dic[1])%mod + (n2[i-1]*dic[0])%mod + (n0[i-1]*dic[2])%mod)%mod

print(n0[n-1])