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].
import sys
input=sys.stdin.readline
n, l, r = map(int, input().split())
zero = r//3-(l-1)//3
one = (r-1)//3-(l-2)//3
two = r-l+1-zero-one
MOD = 10**9+7
dp = [[0]*3 for _ in range(n+1)]
dp[0][0] = 1

for i in range(n):
    dp[i+1][0] = (dp[i][0]*zero%MOD+dp[i][1]*two%MOD+dp[i][2]*one%MOD)%MOD
    dp[i+1][1] = (dp[i][0]*one%MOD+dp[i][1]*zero%MOD+dp[i][2]*two%MOD)%MOD
    dp[i+1][2] = (dp[i][0]*two%MOD+dp[i][1]*one%MOD+dp[i][2]*zero%MOD)%MOD

print(dp[n][0])
