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;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
inline unsigned int R() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
const int N = 233333, mo = 1e9 + 7;
int n, l, r, f[N][3], a[3];
int main() {
  read(n);
  read(l);
  read(r);
  f[0][0] = 1;
  for (register int i = 0; i <= 2; i++) {
    if (r - l <= 100)
      for (register int j = l; j <= r; j++) a[i] += j % 3 == i;
    else {
      int x = l, y = r;
      while (x % 3 != i) x++;
      while (y % 3 != i) y--;
      a[i] = (y - x) / 3 + 1;
    }
  }
  for (register int i = 1; i <= n; i++)
    for (register int j = 0; j <= 2; j++)
      for (register int k = 0; k <= 2; k++) {
        f[i][j] = (f[i][j] + 1LL * f[i - 1][k] * a[(j - k + 3) % 3]) % mo;
      }
  cout << f[n][0];
  return 0;
}
