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].
MOD = pow(10, 9) + 7

n, l, r = map(int, raw_input().strip().split())

a0 = a1 = a2 = 0
xl = l
while xl % 3 != 0 and xl < r: xl += 1
xr = r
while xr % 3 != 0 and xr > xl: xr -= 1

if xl <= xr and xl % 3 == 0 and xr % 3 == 0: a0 = (xr - xl) // 3 + 1

xl = l
while xl % 3 != 1 and xl < r: xl += 1
xr = r
while xr % 3 != 1 and xr > xl: xr -= 1

if xl <= xr and xl % 3 == 1 and xr % 3 == 1: a1 = (xr - xl) // 3 + 1

xl = l
while xl % 3 != 2 and xl < r: xl += 1
xr = r
while xr % 3 != 2 and xr > xl: xr -= 1

if xl <= xr and xl % 3 == 2 and xr % 3 == 2: a2 = (xr - xl) // 3 + 1


dp = [[0 for i in xrange(3)] for j in xrange(n + 1)]

dp[1] = [a0 % MOD, a1 % MOD, a2 % MOD]

for i in xrange(2, n + 1):
    dp[i][0] = dp[i - 1][0] * a0 + dp[i - 1][1] * a2 + dp[i - 1][2] * a1
    if dp[i][0] >= MOD: dp[i][0] %= MOD
    dp[i][1] = dp[i - 1][0] * a1 + dp[i - 1][1] * a0 + dp[i - 1][2] * a2
    if dp[i][1] >= MOD: dp[i][1] %= MOD
    dp[i][2] = dp[i - 1][0] * a2 + dp[i - 1][1] * a1 + dp[i - 1][2] * a0
    if dp[i][2] >= MOD: dp[i][2] %= MOD

print dp[n][0] % MOD