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 int mod = 1e9 + 7;
long long int l, r;
map<long long int, long long int> mp[3];
long long int calc(long long int n, long long int rem) {
  if (rem == 0)
    return n / 3;
  else if (rem == 1)
    return (n + 2) / 3;
  return (n + 1) / 3;
}
long long int fun(long long int n, long long int rem) {
  if (mp[rem].find(n) != mp[rem].end()) return mp[rem][n];
  if (n == 1) {
    long long int ans = calc(r, rem) - calc(l - 1, rem);
    return ans;
  }
  long long int sum = 0;
  sum += fun(n / 2, 0) * fun(n - n / 2, rem) % mod;
  sum %= mod;
  sum += fun(n / 2, 1) * fun(n - n / 2, (3 + rem - 1) % 3) % mod;
  sum %= mod;
  sum += fun(n / 2, 2) * fun(n - n / 2, (3 + rem - 2) % 3) % mod;
  sum %= mod;
  return mp[rem][n] = sum;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n >> l >> r;
  cout << fun(n, 0) << '\n';
  return 0;
}
