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 arr[3];
long long dp[200001][3];
int main() {
  ios::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  long long n, l, r;
  cin >> n >> l >> r;
  if (l == r) {
    if (l % 3 == 1) {
      arr[1] = 1;
    } else if (l % 3 == 2) {
      arr[2] = 1;
    } else {
      arr[0] = 1;
    }
  } else {
    if (l % 3 == 1) {
      l += 2;
      arr[1]++;
      arr[2]++;
    } else if (l % 3 == 2) {
      l += 1;
      arr[2]++;
    }
    if (r % 3 == 1) {
      r -= 1;
      arr[1]++;
    } else if (r % 3 == 2) {
      r -= 2;
      arr[1]++;
      arr[2]++;
    }
    arr[0] += ((r - l) / 3 + 1);
    arr[1] += (arr[0] - 1);
    arr[2] += (arr[0] - 1);
  }
  dp[1][0] = arr[0];
  dp[1][1] = arr[1];
  dp[1][2] = arr[2];
  for (int i = 2; i <= n; i++) {
    dp[i][0] = (dp[i - 1][2] * arr[1]) % (long long)1000000007 +
               (dp[i - 1][1] * arr[2]) % (long long)1000000007 +
               (dp[i - 1][0] * arr[0]) % (long long)1000000007;
    dp[i][1] = (dp[i - 1][0] * arr[1]) % (long long)1000000007 +
               (dp[i - 1][1] * arr[0]) % (long long)1000000007 +
               (dp[i - 1][2] * arr[2]) % (long long)1000000007;
    dp[i][2] = (dp[i - 1][0] * arr[2]) % (long long)1000000007 +
               (dp[i - 1][2] * arr[0]) % (long long)1000000007 +
               (dp[i - 1][1] * arr[1]) % (long long)1000000007;
  }
  cout << dp[n][0] % (long long)1000000007;
  return 0;
}
