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;
int main(void) {
  long long n, i, l, r;
  cin >> n >> l >> r;
  long long nn[3];
  long long dp[n + 1][3];
  for (i = 0; i < 3; i++) nn[i] = (r - i + 3) / 3 - (l - i + 2) / 3;
  for (i = 0; i <= n; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
    dp[i][2] = 0;
  }
  dp[0][0] = 1;
  for (i = 1; i <= n; i++) {
    for (long long j = 0; j <= 2; j++) {
      for (long long k = 0; k <= 2; k++) {
        dp[i][j] += nn[k] * dp[i - 1][(j + 3 - k) % 3];
        dp[i][j] %= 1000000007;
      }
    }
  }
  cout << dp[n][0];
}
