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;
ifstream in("test12.in");
ofstream out("test12.out");
long long mat[3][200005], m[3];
int main() {
  long long n, l, r, i, j, k, temp, order[9] = {0, 2, 1, 1, 0, 2, 2, 1, 0};
  cin >> n >> l >> r;
  for (i = 0; i <= 2; i++) m[(l + i) % 3] = (r + 3 - l - i) / 3;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= 2; j++) {
      if (i == 1)
        mat[j][i] = m[j];
      else {
        for (k = 0; k <= 2; k++) {
          mat[j][i] += mat[k][i - 1] * m[order[j * 3 + k]];
        }
      }
      mat[j][i] %= 1000000007;
    }
  }
  cout << mat[0][n];
  return 0;
}
