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;
const int mod = 1000000007;
int modEqual(int ma, int m) {
  int res = ma / 3;
  if (ma % 3 >= m) res++;
  return res;
}
int main() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long f[3];
  f[0] = modEqual(r, 0) - modEqual(l - 1, 0);
  f[1] = modEqual(r, 1) - modEqual(l - 1, 1);
  f[2] = modEqual(r, 2) - modEqual(l - 1, 2);
  long long fre[] = {1, 0, 0};
  for (long long i = 0; i < n; i++) {
    long long next[3] = {0};
    for (long long a = 0; a < 3; a++)
      for (long long b = 0; b < 3; b++) next[(a + b) % 3] += fre[a] * f[b];
    for (long long k = 0; k < 3; k++) fre[k] = next[k] % mod;
  }
  cout << fre[0] << endl;
  return 0;
}
