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 n, l, r;
long long a[3];
long long b[3];
int main() {
  cin >> n >> l >> r;
  long long a0 = r / 3 - l / 3;
  if (l % 3 == 0) ++a0;
  long long a1 = r / 3 - l / 3;
  if (r != l && r % 3 >= 1) ++a1;
  if (r != l && l % 3 == 2) --a1;
  long long a2 = r - l + 1 - a0 - a1;
  a[0] = a0;
  a[1] = a1;
  a[2] = a2;
  for (int i = 1; i < n; ++i) {
    b[0] = (a0 * a[0] + a[1] * a2 + a[2] * a1) % 1000000007;
    b[1] = (a0 * a[1] + a[0] * a1 + a[2] * a2) % 1000000007;
    b[2] = (a0 * a[2] + a[0] * a2 + a[1] * a1) % 1000000007;
    a[0] = b[0];
    a[1] = b[1];
    a[2] = b[2];
  }
  cout << a[0] << endl;
  return 0;
}
