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 mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (!a) return b;
  return gcd(b % a, a);
}
long long powmod(long long a, long long po) {
  if (po == 0) return 1;
  if (po == 1) return a;
  long long int p1 = powmod(a, po / 2);
  long long int p2 = p1 * p1;
  long long p3 = powmod(a, po % 2);
  return ((p2 % 1000000007) * (p3 % 1000000007)) % 1000000007;
}
long long dp[200005][3];
void solve() {
  int n, l, r;
  cin >> n >> l >> r;
  long long int c1, c2, c3;
  c1 = c2 = c3 = 0;
  c3 = (floor(r / 3.0) - ceil(l / 3.0)) + 1;
  if ((l % 3 == 1 && r % 3 == 0) || (l % 3 != 1 && r % 3 != 0))
    c1 = c3;
  else if (r % 3 == 0)
    c1 = c3 - 1;
  else
    c1 = c3 + 1;
  c2 = r - l + 1 - c3 - c1;
  dp[1][0] = c3;
  dp[1][1] = c1;
  dp[1][2] = c2;
  for (int i = 1; i < n; i++) {
    dp[i + 1][0] = (dp[i][0] * c3 + dp[i][1] * c2 + dp[i][2] * c1) % 1000000007;
    dp[i + 1][1] = (dp[i][0] * c1 + dp[i][1] * c3 + dp[i][2] * c2) % 1000000007;
    dp[i + 1][2] = (dp[i][0] * c2 + dp[i][1] * c1 + dp[i][2] * c3) % 1000000007;
  }
  cout << dp[n][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}
