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 md = 1e9 + 7;
long long n, l, r, k0, k1, k2, ans, dp[200005][3][3];
void up(long long &a, long long b) { a = (a + b) % md; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> r;
  k0 = r / 3 - (l - 1) / 3;
  l++;
  r++;
  k2 = r / 3 - (l - 1) / 3;
  l++;
  r++;
  k1 = r / 3 - (l - 1) / 3;
  dp[0][0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int l = 0; l < 3; ++l) {
      for (int r = 0; r < 3; ++r) {
        up(dp[i][l][r], dp[i - 1][l][r] * k0);
        up(dp[i][l][r], dp[i - 1][(l + 2) % 3][r] * k1);
        up(dp[i][l][r], dp[i - 1][l][(r + 2) % 3] * k2);
      }
    }
  }
  up(ans, dp[n][0][0]);
  up(ans, dp[n][2][2]);
  up(ans, dp[n][1][1]);
  cout << ans;
}
