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(void) {
  int n, l, r;
  cin >> n >> l >> r;
  int arr[3] = {0, 0, 0};
  for (int i = l; i <= min(l + 2, r); ++i) arr[i % 3] = i;
  for (int i = r; i >= max(r - 2, l); --i)
    arr[i % 3] = (i - arr[i % 3]) / 3 + 1;
  vector<vector<long long int> > dp(3, vector<long long int>(n + 1, 0));
  dp[0][1] = arr[0], dp[1][1] = arr[1], dp[2][1] = arr[2];
  for (int i = 2; i <= n; ++i)
    for (int j = 0; j < 3; ++j)
      for (int k = 0; k < 3; ++k) {
        long long int tmp = arr[k] * dp[((j - k) + 3) % 3][i - 1];
        tmp %= 1000000007;
        dp[j][i] += tmp;
        dp[j][i] %= 1000000007;
      }
  cout << dp[0][n] << endl;
  return 0;
}
