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>
const double E = exp(1);
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
using namespace std;
long long dp[3][maxn];
int n, l, r, t;
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin >> n >> l >> r;
  int num0, num1, num2;
  num0 = r / 3 - (l - 1) / 3;
  num1 = r / 3 + (r % 3 >= 2) - (l - 1) / 3 - ((l - 1) % 3 >= 2);
  num2 = r - l + 1 - num0 - num1;
  dp[0][1] = num0;
  dp[1][1] = num1;
  dp[2][1] = num2;
  for (int i = 2; i <= n; i++) {
    dp[0][i] = ((dp[0][i - 1] % mod * (num0 % mod)) % mod +
                (dp[1][i - 1] % mod * num2 % mod) % mod +
                (dp[2][i - 1] % mod * num1 % mod) % mod) %
               mod;
    dp[1][i] = ((dp[0][i - 1] % mod * (num1 % mod)) % mod +
                (dp[1][i - 1] % mod * num0 % mod) % mod +
                (dp[2][i - 1] % mod * num2 % mod) % mod) %
               mod;
    dp[2][i] = ((dp[0][i - 1] % mod * (num2 % mod)) % mod +
                (dp[1][i - 1] % mod * num1 % mod) % mod +
                (dp[2][i - 1] % mod * num0 % mod) % mod) %
               mod;
  }
  cout << dp[0][n] % mod << endl;
  return 0;
}
