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;
const int N = 1e6;
const long long mod = 1e9 + 7;
long long n, l, r, a, b, c;
long long dp[N + 2][4];
long long dfs(int st, int md) {
  if (st > n) {
    if (md == 0)
      return 1;
    else
      return 0;
  }
  if (dp[st][md] != -1) return dp[st][md];
  long long now = 0;
  now = (now + a * dfs(st + 1, md)) % mod;
  now = (now + b * dfs(st + 1, (md + 1) % 3)) % mod;
  now = (now + c * dfs(st + 1, (md + 2) % 3)) % mod;
  return dp[st][md] = now;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  n, l, r;
  cin >> n >> l >> r;
  long long x, y;
  a = r / 3 - (l - 1) / 3;
  x = r / 3;
  if ((r % 3) >= 1) x++;
  y = (l - 1) / 3;
  if ((l - 1) % 3 >= 1) y++;
  b = x - y;
  x = r / 3;
  if (r % 3 >= 2) x++;
  y = (l - 1) / 3;
  if ((l - 1) % 3 >= 2) y++;
  c = x - y;
  memset(dp, -1, sizeof(dp));
  cout << dfs(1, 0) << endl;
  return 0;
}
