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;
const long long M = 1e9 + 7, Maxn = 2e5 + 7;
;
long long n, r, l, t[4], dp[Maxn][4];
int main() {
  cin >> n >> l >> r;
  for (int i = 0; i < 3; i++)
    if (l % 3 == i) {
      t[i] = (r - l + 1) / 3 + (r % 3 != (i + 2) % 3);
      t[(i + 1) % 3] = (r - l + 1) / 3 + (r % 3 == (i + 1) % 3);
      t[(i + 2) % 3] = (r - l + 1) / 3;
    }
  dp[0][0] = t[0], dp[0][1] = t[1], dp[0][2] = t[2];
  for (int i = 1; i < n; i++) {
    dp[i][0] = ((dp[i - 1][0] * t[0]) % M + (dp[i - 1][1] * t[2]) % M +
                (dp[i - 1][2] * t[1]) % M) %
               M;
    dp[i][1] = ((dp[i - 1][0] * t[1]) % M + (dp[i - 1][1] * t[0]) % M +
                (dp[i - 1][2] * t[2]) % M) %
               M;
    dp[i][2] = ((dp[i - 1][0] * t[2]) % M + (dp[i - 1][1] * t[1]) % M +
                (dp[i - 1][2] * t[0]) % M) %
               M;
  }
  cout << dp[n - 1][0];
}
