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
# input=sys.stdin.readline

def fu(l,r):
    return [r//3-(l-1)//3,(r-1)//3-(l-2)//3,(r-2)//3-(l-3)//3]


n,l,r=list(map(int,input().split()))
a=fu(l,r)
m=10**9+7
dp0=[a[0]]
dp1=[a[1]]
dp2=[a[2]]
for i in range(1,n):
    x=(dp0[-1]*a[0]+dp1[-1]*a[2]+dp2[-1]*a[1])%m
    y=(dp0[-1]*a[1]+dp1[-1]*a[0]+dp2[-1]*a[2])%m
    z=(dp0[-1]*a[2]+dp1[-1]*a[1]+dp2[-1]*a[0])%m
    dp0.append(x)
    dp1.append(y)
    dp2.append(z)
print((dp0[-1])%m)