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 BoostIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long const maxn = 100005;
double pi = 3.1415926535897932;
double eps = 1e-7;
long long mod = 1e9 + 7;
long long Pow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    return Pow(a, b / 2) * Pow(a, b / 2);
  } else {
    return Pow(a, b / 2) * Pow(a, b / 2) * a;
  }
}
struct My {
  long long a;
  long long b;
  long long c;
};
int32_t main() {
  BoostIO();
  long long n, l, r;
  cin >> n >> l >> r;
  vector<long long> ost(3);
  while (l <= r && l % 3 != 0) {
    ++ost[l % 3];
    ++l;
  }
  while (r >= l && r % 3 != 2) {
    ++ost[r % 3];
    --r;
  }
  long long k = (r - l + 1) / 3;
  for (long long i = 0; i < 3; ++i) {
    ost[i] += k;
  }
  vector<vector<long long>> dp(n + 1, vector<long long>(3));
  dp[1][0] = ost[0];
  dp[1][1] = ost[1];
  dp[1][2] = ost[2];
  for (long long i = 2; i <= n; ++i) {
    dp[i][0] =
        ((dp[i - 1][0] * ost[0] % mod + dp[i - 1][1] * ost[2] % mod) % mod +
         (dp[i - 1][2] * ost[1]) % mod) %
        mod;
    dp[i][1] =
        ((dp[i - 1][0] * ost[1] % mod + dp[i - 1][1] * ost[0] % mod) % mod +
         (dp[i - 1][2] * ost[2]) % mod) %
        mod;
    dp[i][2] =
        ((dp[i - 1][0] * ost[2] % mod + dp[i - 1][1] * ost[1] % mod) % mod +
         (dp[i - 1][2] * ost[0]) % mod) %
        mod;
  }
  cout << dp[n][0] << endl;
  return 0;
}
