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() {
  ios_base::sync_with_stdio(false);
  long long int n, l, r;
  cin >> n >> l >> r;
  long long int A[3][n + 1];
  long long int cnt[3] = {0, 0, 0};
  long long int k = l % 3;
  cnt[k % 3] = (r - l) / 3 + 1;
  if (r - l - 1 >= 0) cnt[(k + 1) % 3] = (r - l - 1) / 3 + 1;
  if (r - l - 2 >= 0) cnt[(k + 2) % 3] = (r - l - 2) / 3 + 1;
  A[0][1] = cnt[0];
  A[1][1] = cnt[1];
  A[2][1] = cnt[2];
  for (int j = 2; j <= n; j++) {
    A[0][j] = ((A[0][j - 1] * cnt[0]) % 1000000007 +
               (A[1][j - 1] * cnt[2]) % 1000000007) %
              1000000007;
    A[0][j] = (A[0][j] + (A[2][j - 1] * cnt[1]) % 1000000007) % 1000000007;
    A[1][j] = ((A[1][j - 1] * cnt[0]) % 1000000007 +
               (A[0][j - 1] * cnt[1]) % 1000000007) %
              1000000007;
    A[1][j] = (A[1][j] + (A[2][j - 1] * cnt[2]) % 1000000007) % 1000000007;
    A[2][j] = ((A[2][j - 1] * cnt[0]) % 1000000007 +
               (A[1][j - 1] * cnt[1]) % 1000000007) %
              1000000007;
    A[2][j] = (A[2][j] + (A[0][j - 1] * cnt[2]) % 1000000007) % 1000000007;
  }
  cout << A[0][n] << endl;
}
