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 ans[2000100][5], n, mod[5];
long long count(long long x, long long mod3) {
  if (x == n) {
    if (!mod3)
      return 1;
    else
      return 0;
  }
  if (ans[x][mod3] == -1) {
    long long val = 0;
    long long tmp = 1e9 + 7;
    for (int i = 0; i < 3; i++) {
      if (mod[i]) val = (val + mod[i] * count(x + 1, (mod3 + i) % 3)) % tmp;
    }
    ans[x][mod3] = val;
  }
  return ans[x][mod3];
}
int main() {
  memset(ans, -1, sizeof(ans));
  long long l, r;
  cin >> n >> l >> r;
  for (int i = 0; i < 3; i++) {
    mod[i] = (r - l + 1) / 3;
  }
  for (int i = 0; i < (r - l + 1) % 3; i++) {
    long long tmp = l + i;
    mod[tmp % 3]++;
  }
  long long tmp = 1e9 + 7;
  long long ans = count(0, 0) % tmp;
  cout << ans << "\n";
}
