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 power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, l, r;
  cin >> n >> l >> r;
  long long m = 1e9 + 7;
  long long dp[n + 1][3], x = (r - l + 1) / 3, y = (r - l + 1) / 3,
                          z = (r - l + 1) / 3;
  if ((r - l + 1) % 3 == 1) {
    if (l % 3 == 0)
      x++;
    else if (l % 3 == 1)
      y++;
    else
      z++;
  } else if ((r - l + 1) % 3 == 2) {
    if (l % 3 == 0) {
      x++;
      y++;
    } else if (l % 3 == 1) {
      y++;
      z++;
    } else {
      z++;
      x++;
    }
  }
  dp[1][0] = x;
  dp[1][1] = y;
  dp[1][2] = z;
  for (long long i = 2; i <= n; i++) {
    dp[i][0] =
        (dp[i - 1][1] * z % m + dp[i - 1][2] * y % m + dp[i - 1][0] * x % m) %
        m;
    dp[i][1] =
        (dp[i - 1][0] * y % m + dp[i - 1][1] * x % m + dp[i - 1][2] * z % m) %
        m;
    dp[i][2] =
        (dp[i - 1][2] * x % m + dp[i - 1][1] * y % m + dp[i - 1][0] * z % m) %
        m;
  }
  cout << dp[n][0];
  return 0;
}
