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 math
n,l,r=map(int,input().split())
b=[0,0,0]
b[l%3]=(r-l+3)//3
b[(l+1)%3]=(r-l+2)//3
b[(l+2)%3]=(r-l+1)//3

mod=(10**9)+7

dp=[[0 for i in range(3)] for j in range(n+1)]
dp[0][0]=1
dp[0][1]=0
dp[0][2]=0
for i in range(1,n+1):
    dp[i][0]=((b[0]*dp[i-1][0])%mod + (b[1]*dp[i-1][2])%mod + (b[2]*dp[i-1][1])%mod)%mod
    dp[i][1] = ((b[0] * dp[i - 1][1]) % mod + (b[1] * dp[i - 1][0]) % mod + (b[2] * dp[i - 1][2]) % mod) % mod
    dp[i][2]=((b[0]*dp[i-1][2])%mod + (b[1]*dp[i-1][1])%mod + (b[2]*dp[i-1][0])%mod)%mod
print((dp[n][0])%mod)