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;
void printprecise(double l, long long int precision) {
  std::cout << std::fixed;
  std::cout << std::setprecision(precision);
  std::cout << l;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, l, r;
  cin >> n >> l >> r;
  vector<vector<long long int> > v(3, vector<long long int>(3));
  v[0][0] = (r / 3) - (l - 1) / 3;
  v[0][1] = v[0][0] + (l % 3 == 1) - (r % 3 == 0);
  v[0][2] = v[0][0] + (l % 3 != 0) - (r % 3 != 2);
  for (long long int i = 0; i < 3; i++) {
    v[2][i] = v[0][i];
  }
  for (long long int i = 2; i < n + 1; i++) {
    for (long long int j = 0; j < 3; j++) {
      v[1][j] = v[2][j];
      v[2][j] = 0;
    }
    for (long long int j = 0; j < 3; j++) {
      for (long long int k = 0; k < 3; k++) {
        v[2][(k + j) % 3] =
            (v[2][(k + j) % 3] + (v[0][j] * v[1][k]) % 1000000007) % 1000000007;
      }
    }
  }
  cout << v[2][0];
  return 0;
}
