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].
modulo = int(1e9) + 7

n,l,r = map(int, raw_input().split())
cnt = [0 for i in xrange(3)]
for i in xrange(3):
	last = -1
	first = 0
	for j in xrange(l,min(l+3,r+1)):
		if j%3 == i:
			first = j
			break
	for j in xrange(r,max(l-1,r-3),-1):
		if j%3 == i:
			last = j
			break
	if last >= first:
		cnt[i] = (last-first)/3 + 1

	
dp = [[0 for i in xrange(3)] for j in xrange(n+1)]
for i in xrange(3):
	dp[1][i] = cnt[i]
for i in xrange(2,n+1):
	dp[i][0] = (dp[i-1][0]*cnt[0] + dp[i-1][1]*cnt[2] + dp[i-1][2]*cnt[1])%modulo;
	dp[i][1] = (dp[i-1][0]*cnt[1] + dp[i-1][1]*cnt[0] + dp[i-1][2]*cnt[2])%modulo;
	dp[i][2] = (dp[i-1][0]*cnt[2] + dp[i-1][2]*cnt[0] + dp[i-1][1]*cnt[1])%modulo;
print dp[n][0]