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].
# 23:11
import sys
# sys.stdin = open("c.in")
MOD = int(1e9+7)

# dp[i][j] = \sum(k) dp[i-1][j-k]

import time
def solve(n, l, r):
    dp = [[0]*3 for _ in range(n)]

    add_num = [r//3-(l-1)//3, (r+2)//3-(l+2-1)//3, (r+1)//3-(l+1-1)//3]
    add_num = [i % MOD for i in add_num]

    for k in range(3):
        dp[0][k] += add_num[k]

    for i in range(1, n):
        for j in range(3):
            for k in range(3):
                dp[i][j] += (dp[i-1][(j-k+3)%3] * add_num[k])%MOD
                dp[i][j] %= MOD
    print(dp[n-1][0])

if __name__ == '__main__':
    while True:
        try:
            n, l, r = [int(x) for x in input().split()]
            solve(n, l, r)
        except EOFError:
            break