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,raw_input().split())
a = [[0,0,0]for i in xrange(n+1)]
r = r+1
p=[r/3,r/3,r/3]
if r%3==1:
    p[0] += 1
elif r%3== 2:
    p[0]+= 1
    p[1] += 1
q = [l/3,l/3,l/3]
if l%3==1:
    q[0] += 1
elif l%3== 2:
    q[0] += 1
    q[1] += 1
a[1][0] = p[0]-q[0]
a[1][1] = p[1]-q[1]
a[1][2] = p[2]-q[2]
for i in xrange(2,n+1):
    a[i][0] = (a[i-1][0]*a[1][0] + a[i-1][1]*a[1][2] + a[i-1][2]*a[1][1])%1000000007
    a[i][1] = (a[i-1][0]*a[1][1] + a[i-1][1]*a[1][0] + a[i-1][2]*a[1][2])%1000000007
    a[i][2] = (a[i-1][0]*a[1][2] + a[i-1][1]*a[1][1] + a[i-1][2]*a[1][0])%1000000007
print (a[n][0])%1000000007