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;
int main() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long z, o, t;
  z = (r - l + 1) / 3;
  o = z;
  t = z;
  long long dub = r;
  for (int i = 0; i < (r - l + 1) % 3; i++) {
    if (dub % 3 == 0)
      z++;
    else if (dub % 3 == 1)
      o++;
    else
      t++;
    dub--;
  }
  vector<vector<long long>> ans(n, vector<long long>(3));
  ans[0][0] = z;
  ans[0][1] = o;
  ans[0][2] = t;
  long long mod = 1000000007;
  for (long long i = 1; i < n; i++) {
    ans[i][0] = ((ans[i - 1][0] * z) % mod + (ans[i - 1][1] * t) % mod +
                 (ans[i - 1][2] * o) % mod) %
                mod;
    ans[i][1] = ((ans[i - 1][0] * o) % mod + (ans[i - 1][1] * z) % mod +
                 (ans[i - 1][2] * t) % mod) %
                mod;
    ans[i][2] = ((ans[i - 1][0] * t) % mod + (ans[i - 1][1] * o) % mod +
                 (ans[i - 1][2] * z) % mod) %
                mod;
  }
  cout << ans[n - 1][0];
  return 0;
}
