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 int **board;
long long int K[3];
long long int dp(long long int i, long long int j) {
  if (i == 1) {
    if (j == 0) return K[0];
    if (j == 1) return K[1];
    if (j == 2) return K[2];
  }
  for (int k = 0; k < 3; k++) {
    if (board[i - 1][k] == 0) {
      board[i - 1][k] = dp(i - 1, k) % 1000000007;
    }
  }
  long long int ans = 0;
  long long int carry = 0;
  for (int k = 0; k < 3; k++) {
    carry = j - k;
    if (carry < 0) carry += 3;
    ans += ((board[i - 1][carry] % 1000000007) * K[k]) % 1000000007;
    ans = ans % 1000000007;
  }
  return ans;
}
int main() {
  long n, l, r;
  cin >> n >> l >> r;
  board = (long long int **)malloc(n * sizeof(long long int *));
  for (int i = 0; i < n; i++) {
    board[i] = (long long int *)malloc(3 * sizeof(long long int));
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 3; j++) board[i][j] = 0;
  }
  K[0] = int((floor(r / 3.0) - ceil(l / 3.0) + 1)) % 1000000007;
  K[1] = int((floor((r - 1) / 3.0) - ceil((l - 1) / 3.0) + 1)) % 1000000007;
  K[2] = int((floor((r - 2) / 3.0) - ceil((l - 2) / 3.0) + 1)) % 1000000007;
  long long int ans = dp(n, 0);
  cout << ans;
  return 0;
}
