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

def add(x, y):
    global mod
    return (x + y) % mod

def sub(x, y):
    global mod
    return ((x - y) % mod + mod) % mod

def mul(x, y):
    global mod
    return (x * y) % mod

def binpow(x, y):
    global mod
    if y == 0:
        return 1
    elif y % 2 == 0:
        z = binpow(x, y // 2)
        return mul(z, z)
    else:
        return mul(binpow(x, y - 1), x)

def divide(x, y):
    global mod
    return mul(x, binpow(y, mod - 2))

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

ans = [1, 0, 0]
cnt = [r // 3 - (l - 1) // 3, (r + 2) // 3 - (l + 1) // 3, (r + 1) // 3 - l // 3]

for i in range(n):
    newans = [0, 0, 0]
    for j in range(3):
        for k in range(3):
            newans[(j + k) % 3] = add(newans[(j + k) % 3], mul(cnt[k], ans[j]))
    ans = newans

print(ans[0])