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;
const long long int mod = 1e9 + 7, N = 2e5;
long long int dp[3][N + 1];
int main() {
  long long int n, l, r;
  cin >> n >> l >> r;
  long long int A[3];
  bool a = r % 3, b = (l - 1) % 3;
  A[0] = (r / 3 - (l - 1) / 3) % mod;
  A[1] = (r / 3 + a - ((l - 1) / 3 + b)) % mod;
  A[2] = (r / 3 + (r % 3 / 2) - ((l - 1) / 3 + (l - 1) % 3 / 2)) % mod;
  for (int i = 0; i < 3; i++) {
    dp[i][1] = A[i];
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j < 3; j++) {
      if (j == 0) {
        dp[j][i] = (A[0] * dp[0][i - 1] % mod + A[1] * dp[2][i - 1] % mod +
                    A[2] * dp[1][i - 1] % mod) %
                   mod;
      } else if (j == 1) {
        dp[j][i] = (A[2] * dp[2][i - 1] % mod + A[1] * dp[0][i - 1] % mod +
                    A[0] * dp[1][i - 1] % mod) %
                   mod;
      } else {
        dp[j][i] = (A[1] * dp[1][i - 1] % mod + A[2] * dp[0][i - 1] % mod +
                    A[0] * dp[2][i - 1] % mod) %
                   mod;
      }
    }
  }
  cout << dp[0][n];
}
