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 N = 2 * (1e5) + 5;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long i, x, y, z, j, k, m, n, q, l, r, c, pos, ans, num;
  cin >> n >> l >> r;
  vector<long long> v(3, 0);
  if (r - l < 2) {
    while (l <= r) {
      v[l % 3]++;
      l++;
    }
    x = v[0];
    y = v[1];
    z = v[2];
  } else {
    while (l % 3 != 0) {
      v[l % 3]++;
      l++;
    }
    while (r % 3 != 0) {
      v[r % 3]++;
      r--;
    }
    j = (r - l) / 3;
    x = v[0] + 1 + j;
    y = v[1] + j;
    z = v[2] + j;
  }
  vector<vector<long long> > dp(N, vector<long long>(3));
  dp[0][0] = x;
  dp[0][1] = y;
  dp[0][2] = z;
  for (long long(i) = (1); (i) < (n); ++(i)) {
    dp[i][0] =
        ((dp[i - 1][0] * x) % 1000000007 + (dp[i - 1][1] * z) % 1000000007 +
         (dp[i - 1][2] * y) % 1000000007) %
        1000000007;
    dp[i][1] =
        ((dp[i - 1][0] * y) % 1000000007 + (dp[i - 1][1] * x) % 1000000007 +
         (dp[i - 1][2] * z) % 1000000007) %
        1000000007;
    dp[i][2] =
        ((dp[i - 1][0] * z) % 1000000007 + (dp[i - 1][1] * y) % 1000000007 +
         (dp[i - 1][2] * x) % 1000000007) %
        1000000007;
  }
  cout << dp[n - 1][0];
}
