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;
void getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
const int MO = 1e9 + 7;
int n, l, r;
long long ans, f[200010][3], s[3];
int main() {
  cin >> n >> l >> r;
  --l;
  s[0] = r / 3 - l / 3;
  s[1] = (r + 2) / 3 - (l + 2) / 3;
  s[2] = (r + 1) / 3 - (l + 1) / 3;
  f[0][0] = 1;
  for (int i = 0; i <= n - 1; ++i)
    for (int j = 0; j <= 2; ++j)
      if (f[i][j]) {
        for (int k = 0; k <= 2; ++k) {
          (f[i + 1][(k + j) % 3] += f[i][j] * s[k]) %= MO;
        }
      }
  ans = (f[n][0] % MO + MO) % MO;
  cout << ans;
  return 0;
}
