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 int mod = 1e9 + 7;
int main() {
  int n, l, r;
  scanf("%d %d %d", &n, &l, &r);
  vector<vector<long long> > dp(n + 1, vector<long long>(3, 0));
  vector<long long> a(3);
  a[0] = r / 3 - (l - 1) / 3;
  a[1] = (r + 2) / 3 - (l + 1) / 3;
  a[2] = (r + 1) / 3 - l / 3;
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < 3; k++) {
        int x = (j + k) % 3;
        dp[i + 1][x] += (dp[i][j] * a[k]);
        dp[i + 1][x] %= mod;
      }
    }
  }
  printf("%lld\n", dp[n][0]);
}
