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[200005][3];
long long int n, l, r, md;
long long int a, b, c;
int main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> l >> r;
  a = r / 3 - (l - 1) / 3;
  b = c = max(0LL, a - 1);
  long long int last = 0;
  for (long long int i = l; i <= r; i++) {
    last = max(last, i);
    if (i % 3 == 0) break;
    if (i % 3 == 1)
      b++;
    else
      c++;
  }
  for (long long int i = r; i > last; i--) {
    if (i % 3 == 0) break;
    if (i % 3 == 1)
      b++;
    else
      c++;
  }
  md = 1e9 + 7;
  dp[0][0] = a;
  dp[0][1] = b;
  dp[0][2] = c;
  for (long long int i = 1; i < n; i++) {
    dp[i][0] =
        ((dp[0][0] * dp[i - 1][0]) % md + (dp[0][1] * dp[i - 1][2]) % md +
         (dp[0][2] * dp[i - 1][1]) % md) %
        md;
    dp[i][1] =
        ((dp[0][0] * dp[i - 1][1]) % md + (dp[0][1] * dp[i - 1][0]) % md +
         (dp[0][2] * dp[i - 1][2]) % md) %
        md;
    dp[i][2] =
        ((dp[0][0] * dp[i - 1][2]) % md + (dp[0][1] * dp[i - 1][1]) % md +
         (dp[0][2] * dp[i - 1][0]) % md) %
        md;
  }
  cout << dp[n - 1][0] << endl;
}
