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())


dp = [];
for i in range(0,n):
    dp.append([0,0,0]);

dp[0][0] = (r-0)//3-(l-0-1)//3
dp[0][1] = (r-1)//3-(l-1-1)//3
dp[0][2] = (r-2)//3-(l-2-1)//3

#print(dp)

MOD = int(1e9 + 7);

for i in range(1 , n):
    dp[i][0] += dp[i-1][0]*dp[0][0] + dp[i-1][1]*dp[0][2] + dp[i-1][2]*dp[0][1];
    dp[i][1] += dp[i-1][0]*dp[0][1] + dp[i-1][1]*dp[0][0] + dp[i-1][2]*dp[0][2];
    dp[i][2] += dp[i-1][0]*dp[0][2] + dp[i-1][1]*dp[0][1] + dp[i-1][2]*dp[0][0];
    dp[i][0]%=MOD;
    dp[i][1]%=MOD;
    dp[i][2]%=MOD;



print(dp[n-1][0])
