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 maxn = 2e5 + 100;
const int mod = 1e9 + 7;
long long dp[maxn][3];
long long n, l, r;
int main() {
  cin >> n >> l >> r;
  int a = r / 3 - (l - 1) / 3;
  int b = (r + 2) / 3 - (l + 1) / 3;
  int c = (r + 1) / 3 - l / 3;
  dp[1][0] = a, dp[1][1] = b, dp[1][2] = c;
  for (int i = 2; i <= n; i++) {
    dp[i][0] = (dp[i - 1][0] * a % mod + dp[i - 1][1] * c % mod +
                dp[i - 1][2] * b % mod) %
               mod;
    dp[i][1] = (dp[i - 1][0] * b % mod + dp[i - 1][1] * a % mod +
                dp[i - 1][2] * c % mod) %
               mod;
    dp[i][2] = (dp[i - 1][0] * c % mod + dp[i - 1][1] * b % mod +
                dp[i - 1][2] * a % mod) %
               mod;
  }
  cout << dp[n][0];
  return 0;
}
