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;
long long get(long long n) {
  long long ans = n / 3;
  if (n % 3 == 1 || n % 3 == 2) ans++;
  return ans;
}
long long get3(long long n) {
  long long ans = n / 3;
  if (n % 3 == 2) ans++;
  return ans;
}
const int mod = 1e9 + 7;
const int N = 2e5 + 9;
long long dp[N][3], n, t, t2, t3;
long long yo(long long i, long long m) {
  if (i == n + 1) {
    if (m == 0)
      return 1;
    else
      return 0;
  }
  long long &ret = dp[i][m];
  if (ret != -1)
    return ret;
  else
    return ret = ((t * yo(i + 1, m) % mod + t2 * yo(i + 1, (m + 1) % 3) % mod) %
                      mod +
                  t3 * yo(i + 1, (m + 2) % 3) % mod) %
                 mod;
}
int main() {
  long long i, j, k, l, r, m;
  cin >> n >> l >> r;
  t = r / 3 - (l - 1) / 3;
  t2 = get(r) - get(l - 1);
  t3 = get3(r) - get3(l - 1);
  memset(dp, -1, sizeof dp);
  cout << yo(1, 0) << endl;
  return 0;
}
