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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
signed main(void) {
  long long n, l, r;
  cin >> n >> l >> r;
  vector<long long> a(3, r / 3);
  for (long long i = 0; i < (3); i++)
    if (i < r % 3) a[(i + 1) % 3]++;
  l--;
  vector<long long> b(3, l / 3);
  for (long long i = 0; i < (3); i++)
    if (i < l % 3) b[(i + 1) % 3]++;
  for (long long i = 0; i < (3); i++) a[i] -= b[i];
  vector<vector<long long> > dp(n, vector<long long>(3, 0));
  for (long long i = 0; i < (3); i++) dp[0][i] = a[i];
  for (long long i = 0; i < (n - 1); i++)
    for (long long j = 0; j < (3); j++)
      if (dp[i][j]) {
        for (long long k = 0; k < (3); k++) {
          long long t = dp[i][j] * a[k] % 1000000007;
          dp[i + 1][(j + k) % 3] = (dp[i + 1][(j + k) % 3] + t) % 1000000007;
        }
      }
  cout << dp[n - 1][0] << endl;
}
