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;
long long dp[200005][3], a[3], n, l, r;
long long cal(long long mx, long long rem) {
  if (mx < rem) return 0;
  long long dif = mx - rem;
  dif /= 3;
  return dif + 1;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> n >> l >> r;
  for (int i = 0; i < 3; ++i) {
    a[i] = (cal(r, i) - cal(l - 1, i) + 1000000007ll) % 1000000007ll;
  }
  dp[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 3; ++j) {
      for (int k = 0; k < 3; ++k) {
        dp[i][(j + k) % 3] =
            (dp[i][(j + k) % 3] + (a[k] * dp[i - 1][j]) % 1000000007ll) %
            1000000007ll;
      }
    }
  }
  cout << dp[n][0] << '\n';
  return 0;
}
