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].
from sys import stdin
from copy import deepcopy

n, l, r = map(int, stdin.readline().split())
l -= 1
div1, div2, mod = l // 3, r // 3, 1000000007
all = [div2, div2 + 1 if r % 3 else div2, div2 + (min(r % 3, 2) // 2)]
minus = [div1, div1 + 1 if l % 3 else div1, div1 + (min(l % 3, 2) // 2)]
all = [all[i] - minus[i] for i in range(3)]

mem, p = [deepcopy(all), [0, 0, 0]], 0

for i in range(1, n):
    p ^= 1
    for j in range(1, 4):
        for k in range(1, 4):
            tem = (j + k) % 3
            mem[p][tem] = mem[p][tem] % mod + (mem[p ^ 1][j % 3] * all[k % 3]) % mod

    for j in range(3):
        mem[p ^ 1][j] = 0

print(mem[p][0] % mod)
