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())

mod = 10**9 + 7
dp = [[0 for j in range(3)] for _ in range(n+10)]

dp[0][0] = 1

m0 = r // 3
m1 = r // 3
m2 = r // 3

if r % 3 == 1:
    m1 += 1
elif r % 3 == 2:
    m1 += 1
    m2 += 1

l -= 1
m0 -= l // 3
m1 -= l // 3
m2 -= l // 3

if l % 3 == 1:
    m1 -= 1
elif l % 3 == 2:
    m1 -= 1
    m2 -= 1

for i in range(n):
    dp[i+1][0] += dp[i][0] * m0
    dp[i+1][1] += dp[i][0] * m1
    dp[i+1][2] += dp[i][0] * m2

    dp[i + 1][1] += dp[i][1] * m0
    dp[i + 1][2] += dp[i][1] * m1
    dp[i + 1][0] += dp[i][1] * m2

    dp[i + 1][2] += dp[i][2] * m0
    dp[i + 1][0] += dp[i][2] * m1
    dp[i + 1][1] += dp[i][2] * m2

    dp[i+1][0] %= mod
    dp[i + 1][1] %= mod
    dp[i + 1][2] %= mod

print(dp[n][0])
