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 mod = 1e9 + 7;
const long long inf = 1e18;
void problem_solver() {
  long long n, l, r;
  cin >> n >> l >> r;
  vector<long long> rem(3, 0);
  rem[0] = r / 3 - (l + 2) / 3 + 1;
  rem[1] = rem[0] - 1;
  rem[2] = rem[1];
  if (l % 3 == 1) {
    rem[1] += 1;
    rem[2] += 1;
  }
  if (l % 3 == 2) {
    rem[2] += 1;
  }
  if (r % 3 == 2) {
    rem[1] += 1;
    rem[2] += 1;
  }
  if (r % 3 == 1) {
    rem[1] += 1;
  }
  vector<vector<long long>> dp(n + 1, vector<long long>(3, 0));
  dp[1][0] = rem[0];
  dp[1][1] = rem[1];
  dp[1][2] = rem[2];
  void(0);
  ;
  for (long long i = 2; i <= n; i++) {
    dp[i][0] = (dp[i - 1][0] * rem[0]) % mod + (dp[i - 1][1] * rem[2]) % mod +
               (dp[i - 1][2] * rem[1]) % mod;
    dp[i][1] = (dp[i - 1][0] * rem[1]) % mod + (dp[i - 1][1] * rem[0]) % mod +
               (dp[i - 1][2] * rem[2]) % mod;
    dp[i][2] = (dp[i - 1][0] * rem[2]) % mod + (dp[i - 1][1] * rem[1]) % mod +
               (dp[i - 1][2] * rem[0]) % mod;
  }
  cout << dp[n][0] % mod;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) {
    problem_solver();
    cout << "\n";
  }
}
