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 int N = 1e6 + 6;
const int INF = 1e9 + 10;
const long double Pi = 3.141592653589793116, EPS = 1e-9,
                  Gold = ((1 + sqrt(5)) / 2);
map<int, int> lua;
long long f[N][3], n, l, r, A, B, C;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> r;
  A = r / 3 - (l - 1) / 3;
  B = (r + 1) / 3 - (l - 1 + 1) / 3;
  C = (r + 2) / 3 - (l - 1 + 2) / 3;
  f[1][0] = A;
  f[1][1] = B;
  f[1][2] = C;
  for (int i = 2; i <= n; i++) {
    f[i][1] +=
        (f[i - 1][0] * B % 1000000007ll + f[i - 1][1] * A % 1000000007ll +
         f[i - 1][2] * C % 1000000007ll) %
        1000000007ll;
    f[i][0] +=
        (f[i - 1][0] * A % 1000000007ll + f[i - 1][1] * C % 1000000007ll +
         f[i - 1][2] * B % 1000000007ll) %
        1000000007ll;
    f[i][2] +=
        (f[i - 1][0] * C % 1000000007ll + f[i - 1][1] * B % 1000000007ll +
         f[i - 1][2] * A % 1000000007ll) %
        1000000007ll;
  }
  cout << f[n][0];
  return 0;
}
