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].
m = 10**9+7

n, l, r = (int(x) for x in input().split())

t = r-l+1
tm = None
if (r+1)%3==l%3:
	tm = (t//3,)*3
elif ((r+1)%3 - l%3)%3==1:
	a=[0,0,0]
	a[l%3] = 1
	tm = ( t//3+a[0], t//3+a[1], t//3+a[2] )
else:
	a=[0,0,0]
	a[l%3] = 1
	a[r%3] = 1
	tm = ( t//3+a[0], t//3+a[1], t//3+a[2] )

#assert t == tm1+tm2+tm3
#print(tm)

ans0 = ans1 = ans2 = 0 #for n = 0
ans0, ans1, ans2 = tm #for n = 1
ans0o, ans1o, ans2o = tm
for i in range(n-1):
	ans0n = (ans0o*ans0+ans2o*ans1+ans1o*ans2)%m
	ans1n = (ans0o*ans1+ans1o*ans0+ans2o*ans2)%m
	ans2n = (ans0o*ans2+ans1o*ans1+ans2o*ans0)%m 
	ans0o, ans1o, ans2o = ans0n, ans1n, ans2n
	#print(ans0o, ans1o, ans2o )
#ans3 = t**n

print(ans0o%m)