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, input().split())
q = 10 ** 9 + 7
u0 = [0] * n
u1 = [0] * n
u2 = [0] * n
k = (R - L + 1) // 3
c0 = k; c1 = k; c2 = k
if (R - L + 1) % 3 != 0:
    if R % 3 == 0: c0 += 1
    elif R % 3 == 1: c1 += 1
    else: c2 += 1
    if (R - L + 1) % 3 == 2:
        if L % 3 == 0: c0 += 1
        elif L % 3 == 1: c1 += 1
        else: c2 += 1
u0[0] = c0; u1[0] = c1; u2[0] = c2
##print(u0)
##print(u1)
##print(u2)
for i in range(1, n):
    u0[i] = (u0[i-1] * c0 % q + u1[i-1] * c2 % q + u2[i-1] * c1 % q) % q
    u1[i] = (u0[i-1] * c1 % q + u1[i-1] * c0 % q + u2[i-1] * c2 % q) % q
    u2[i] = (u0[i-1] * c2 % q + u1[i-1] * c1 % q + u2[i-1] * c0 % q) % q
##    print(u0)
##    print(u1)
##    print(u2)
print(u0[n - 1])
