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 int dp[1100005][3];
long long int power(long long int x, long long int y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int l, r, n;
  cin >> n >> l >> r;
  long long int a, b, c;
  a = r / 3 - l / 3;
  if (l % 3 == 0) a++;
  long long int p = r - l + 1 - a;
  if (p % 2 == 0)
    b = c = p / 2;
  else {
    if (l % 3 == 1)
      b = p - p / 2, c = p / 2;
    else
      c = p - p / 2, b = p / 2;
  }
  dp[0][0] = a;
  dp[0][1] = b;
  dp[0][2] = c;
  for (int i = 1; i <= n - 1; i++) {
    dp[i][0] =
        ((dp[i - 1][0] * a) % 1000000007 + (dp[i - 1][1] * c) % 1000000007 +
         (dp[i - 1][2] * b) % 1000000007) %
        1000000007;
    dp[i][1] =
        ((dp[i - 1][0] * b) % 1000000007 + (dp[i - 1][1] * a) % 1000000007 +
         (dp[i - 1][2] * c) % 1000000007) %
        1000000007;
    dp[i][2] =
        ((dp[i - 1][0] * c) % 1000000007 + (dp[i - 1][1] * b) % 1000000007 +
         (dp[i - 1][2] * a) % 1000000007) %
        1000000007;
  }
  cout << dp[n - 1][0] << endl;
  return 0;
}
