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].
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, l, r;
  cin >> n >> l >> r;
  r++;
  long long d_0 = (r + ((3 - (r % 3)) % 3) - l) / 3;
  r++, l++;
  long long d_2 = (r + ((3 - (r % 3)) % 3) - l) / 3;
  r++, l++;
  long long d_1 = (r + ((3 - (r % 3)) % 3) - l) / 3;
  vector<vector<long long>> dp(n, vector<long long>(3, 0));
  dp[0][0] = d_0;
  dp[0][1] = d_1;
  dp[0][2] = d_2;
  for (long long i = 1; i < n; ++i) {
    dp[i][0] =
        (dp[i - 1][0] * d_0 % 1000000007 + dp[i - 1][1] * d_2 % 1000000007 +
         dp[i - 1][2] * d_1 % 1000000007) %
        1000000007;
    dp[i][1] =
        (dp[i - 1][0] * d_1 % 1000000007 + dp[i - 1][1] * d_0 % 1000000007 +
         dp[i - 1][2] * d_2 % 1000000007) %
        1000000007;
    dp[i][2] =
        (dp[i - 1][0] * d_2 % 1000000007 + dp[i - 1][1] * d_1 % 1000000007 +
         dp[i - 1][2] * d_0 % 1000000007) %
        1000000007;
  }
  cout << dp[n - 1][0];
  return 0;
}
