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 = (10**9) + 7

n,l,r = list(map(int, input().split()))


def find_ks(l, r):
    num_elems = (r-l+1)
    num_trips = int(num_elems / 3)
    k = [num_trips] * 3

    start_idx = l % 3
    count = num_elems % 3

    for i in range(count):
        k[(start_idx + i)%3] += 1

    return k

k = find_ks(l, r)
f = [[0 for _ in range(3)] for _ in range(n)]


for mod in range(3):
    f[0][mod] = k[mod]

# print(k)
for idx in range(1, n, 1):
    # print(idx)
    for mod in range(3):
        f[idx][mod] += ( f[idx-1][mod] * k[0] ) % MOD
        f[idx][mod] = f[idx][mod] % MOD

        f[idx][mod] += ( f[idx-1][(mod - 1) % 3] * k[1] ) % MOD
        f[idx][mod] = f[idx][mod] % MOD

        f[idx][mod] += ( f[idx-1][(mod - 2) % 3] * k[2] ) % MOD
        f[idx][mod] = f[idx][mod] % MOD

# print(f)

print(f[n-1][0] % MOD)