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;
void solve() {
  long long int n, l, r;
  cin >> n >> l >> r;
  long long int no = r - l + 1;
  long long int nums[3] = {no / 3, no / 3, no / 3};
  if (no % 3 == 1) {
    nums[l % 3]++;
  } else if (no % 3 == 2) {
    nums[l % 3]++;
    nums[(l + 1) % 3]++;
  }
  long long int dp[200005][3];
  for (long long int i = 0; i < 3; i++) {
    dp[0][i] = nums[i] % 1000000007;
  }
  for (long long int i = 1; i < n; i++) {
    dp[i][0] = ((dp[i - 1][0] * nums[0]) % 1000000007 +
                (dp[i - 1][1] * nums[2]) % 1000000007 +
                (dp[i - 1][2] * nums[1]) % 1000000007) %
               1000000007;
    dp[i][1] = ((dp[i - 1][0] * nums[1]) % 1000000007 +
                (dp[i - 1][1] * nums[0]) % 1000000007 +
                (dp[i - 1][2] * nums[2]) % 1000000007) %
               1000000007;
    dp[i][2] = ((dp[i - 1][0] * nums[2]) % 1000000007 +
                (dp[i - 1][1] * nums[1]) % 1000000007 +
                (dp[i - 1][2] * nums[0]) % 1000000007) %
               1000000007;
  }
  cout << dp[n - 1][0] << "\n";
}
int32_t main() {
  long long int t = 1;
  for (long long int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
