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;
long long int solve(long long int i, long long int val) {
  if (i == n) {
    if (val) return 0;
    return 1;
  }
  if (dp[i][val] != -1) return dp[i][val];
  long long int ans = 0;
  for (long long int ii = 0; ii < 3; ii++) {
    long long int z = 0;
    if (ii == 0) {
      if (val == 0)
        z = a;
      else if (val == 1)
        z = c;
      else if (val == 2)
        z = b;
    } else if (ii == 1) {
      if (val == 0)
        z = b;
      else if (val == 1)
        z = a;
      else if (val == 2)
        z = c;
    } else {
      if (val == 0)
        z = c;
      else if (val == 1)
        z = b;
      else if (val == 2)
        z = a;
    }
    ans = (ans + (solve(i + 1, ii) * z) % md) % md;
  }
  return dp[i][val] = ans;
}
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;
  cout << solve(0, 0) << endl;
}
