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().strip().split())
div = 3
mod = 1000000000+7

dp = [[0 for j in range(div)] for i in range(n)]

NUM = [0] * div

for i in range(div):
    NUM[i] += (r-l+1)/div

res = (r-l+1) - (r-l+1)/div*div

if res == 0:
    pass
elif res == 1:
    NUM[r%3] += 1
else: #2
    NUM[r%3] += 1
    NUM[(r-1)%3] += 1

for i in range(div):
    dp[0][i] = NUM[i]%mod

for i in range(1, n):
    for j in range(div):
        for k in range(div):
            dp[i][(j+k)%div] += (dp[i-1][j]*NUM[k])%mod
            dp[i][(j+k)%div] %= mod

print dp[n-1][0]