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].
def mod(x):
    return x%1000_000_007

n, l, r=map(int, input().split())

res=0

if l==r:
    if (l*n)%3==0: res=1
    print(res)
    exit()

if n==1:
    while l%3>0: l=l+1
    if r-l>=0: res=1+((r-l)//3)
    print(res)
    exit()

x=[0,0,0]

x[l%3]=max(0,1+(r-l)//3)
x[(l+1)%3]=max(0,1+(r-l-1)//3)
x[(l+2)%3]=max(0,1+(r-l-2)//3)

res=[x[0], x[1], x[2]]

for i in range(1, n):
    a0=res[0]*x[0]+res[1]*x[2]+res[2]*x[1]
    a1=res[0]*x[1]+res[1]*x[0]+res[2]*x[2]
    a2=res[0]*x[2]+res[1]*x[1]+res[2]*x[0]
    res=[mod(a0), mod(a1), mod(a2)]
    # print(res)

print(res[0])

