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().strip().split())
dp = [[0 for x in range(3)] for j in range(n+1)]
start = l % 3;
mod = 10**9 + 7;
if start == 0:
    one = (r - l)// 3 + 1
    two = (r -l - 1)//3 + 1 if r >= l +1 else 0 ;
    three = (r - l - 2) // 3 + 1 if r >= l + 2 else 0;
elif start == 1:
    two = (r-l) // 3 + 1;
    three = (r - l - 1) // 3 + 1 if r >= l + 1 else 0;
    one = (r - l - 2) // 3 + 1 if r >= l + 2 else 0;
else:
    three= (r - l) // 3 + 1;
    one = (r - l - 1) // 3 + 1 if r >= l + 1 else 0;
    two = (r - l - 2) // 3 + 1 if r >= l + 2 else 0;
for x in range(3):
    dp[1][0] = one
    dp[1][1] = two
    dp[1][2] = three
for x in range(2,n+1):
        dp[x][0] =( dp[x-1][0] * dp[1][0] + dp[x-1][1] * dp[1][2] + dp[x-1][2] * dp[1][1])  % mod;
        dp[x][1] = (dp[x-1][0] * dp[1][1] + dp[x-1][1] * dp[1][0] + dp[x-1][2] * dp[1][2]) % mod;
        dp[x][2] = (dp[x-1][0] * dp[1][2] + dp[x-1][1] * dp[1][1] + dp[x-1][2] * dp[1][0]) % mod;
print(dp[n][0])

