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() {
  int n;
  long long l, r;
  cin >> n >> l >> r;
  long long m3[3] = {0};
  m3[0] = r / 3 - (l - 1) / 3;
  if (l > 1)
    m3[1] = (r - 1) / 3 - (l - 2) / 3;
  else
    m3[1] = (r - 1) / 3 + 1;
  if (l > 2)
    m3[2] = (r - 2) / 3 - (l - 3) / 3;
  else if (r >= 2)
    m3[2] = m3[2] = (r - 2) / 3 + 1;
  else
    m3[2] = 0;
  long long dp[3][n + 1], p = 1e9 + 7;
  dp[0][1] = m3[0];
  dp[1][1] = m3[1];
  dp[2][1] = m3[2];
  for (int i = 2; i < n + 1; i++) {
    dp[0][i] = dp[1][i] = dp[2][i] = 0;
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < 3; k++) {
        dp[(j + k) % 3][i] = (dp[(j + k) % 3][i] + dp[j][i - 1] * m3[k]) % p;
      }
    }
  }
  cout << dp[0][n] % p << endl;
  return 0;
}
