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].
# cf 1105 C (1500)
n, l, r = map(int, input().split())

div0, div1, div2 = 0, 0, 0
while l <= r:
    if l % 3 == 0:
        div0 += 1
        l += 1
    elif l % 3 == 2:
        div2 += 1
        l += 1

    if l <= r:
        if r % 3 == 1:
            div1 += 1
            r -= 1
        elif r % 3 == 2:
            div2 += 1
            r -= 1
    if l < r and l % 3 == 1 and r % 3 == 0:
        a = (r - l + 1) // 3
        div0 += a
        div1 += a
        div2 += a
        break

MOD = int(1e9) + 7
d0, d1, d2 = div0, div1, div2
for i in range(n - 1):
    d0_ = (d0 * div0) + (d1 * div2) + (d2 * div1)
    d1_ = (d0 * div1) + (d1 * div0) + (d2 * div2)
    d2_ = (d0 * div2) + (d1 * div1) + (d2 * div0)
    d0, d1, d2 = d0_ % MOD, d1_ % MOD, d2_ % MOD
print(d0)

