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 = 1000000007;
long long dp[200007][3];
long long n, l, r;
unordered_map<long long, long long> m;
long long solve(long long ind, long long sum) {
  if (ind == n) {
    if (sum == 0)
      return 1;
    else
      return 0;
  }
  if (dp[ind][sum] != -1) return dp[ind][sum];
  long long res = 0;
  for (long long i = 0; i <= 2; i++)
    res = (res % mod +
           ((solve(ind + 1, (sum + i) % 3) % mod) * (m[i] % mod)) % mod) %
          mod;
  return dp[ind][sum] = res % mod;
}
int main() {
  cin >> n >> l >> r;
  m[l % 3] = (r - l);
  m[(l + 1) % 3] = (r - l - 1);
  m[(l + 2) % 3] = (r - l - 2);
  for (int i = 0; i <= 2; i++) {
    if (m[i] >= 0)
      m[i] = m[i] / 3 + 1;
    else
      m[i] = 0;
  }
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0) << endl;
  return 0;
}
