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;
  vector<vector<long long>> a(n + 1, vector<long long>(3, 0));
  long long nums[3] = {0, 0, 0};
  long long l3, r3;
  if (l % 3 == 1) {
    nums[1]++;
    nums[2]++;
    l3 = (l + 2) / 3;
  } else if (l % 3 == 2) {
    nums[2]++;
    l3 = (l + 1) / 3;
  } else {
    l3 = l / 3;
  }
  if (r % 3 == 1) {
    nums[1]++;
    r3 = (r - 1) / 3;
  } else if (r % 3 == 2) {
    nums[1]++;
    nums[2]++;
    r3 = (r - 2) / 3;
  } else {
    r3 = r / 3;
  }
  nums[0] = r3 - l3 + 1;
  nums[1] += (r3 - l3);
  nums[2] += (r3 - l3);
  long long md = 1000000007;
  a[n][0] = nums[0];
  a[n][1] = nums[1];
  a[n][2] = nums[2];
  for (int i = n - 1; i >= 1; i--) {
    a[i][0] = ((nums[0] * a[i + 1][0]) % md + (nums[1] * a[i + 1][2]) % md +
               (nums[2] * a[i + 1][1]) % md) %
              md;
    a[i][1] = ((nums[1] * a[i + 1][0]) % md + (nums[2] * a[i + 1][2]) % md +
               (nums[0] * a[i + 1][1]) % md) %
              md;
    a[i][2] = ((nums[2] * a[i + 1][0]) % md + (nums[0] * a[i + 1][2]) % md +
               (nums[1] * a[i + 1][1]) % md) %
              md;
  }
  cout << a[1][0] << endl;
  return 0;
}
