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 long long mod = 1e9 + 7;
long long N, l, r, dp[200005][3];
void countt(long long &a, long long &b, long long &c, long long x) {
  a = b = c = 0;
  while (x % 3) {
    if (x % 3 == 1)
      b++;
    else
      c++;
    x--;
  }
  a = x / 3 + 1;
  b += x / 3;
  c += x / 3;
}
int main() {
  long long a, b, c;
  cin >> N >> l >> r;
  long long da, db, dc;
  countt(da, db, dc, l - 1);
  countt(a, b, c, r);
  a -= da, b -= db, c -= dc;
  dp[1][0] = a, dp[1][1] = b, dp[1][2] = c;
  for (int i = 2; i <= N; i++) {
    dp[i][0] = (((dp[i - 1][0] * a) % mod + (dp[i - 1][1] * c) % mod) % mod +
                (dp[i - 1][2] * b) % mod) %
               mod;
    dp[i][1] = (((dp[i - 1][0] * b) % mod + (dp[i - 1][1] * a) % mod) % mod +
                (dp[i - 1][2] * c) % mod) %
               mod;
    dp[i][2] = (((dp[i - 1][0] * c) % mod + (dp[i - 1][1] * b) % mod) % mod +
                (dp[i - 1][2] * a) % mod) %
               mod;
  }
  cout << dp[N][0];
  return 0;
}
