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;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % MOD;
    a = a * a % MOD;
    b /= 2;
  }
  return ans;
}
long long a[44];
long long dp[4][200001], mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  int n, l, r;
  cin >> n >> l >> r;
  for (int i = l; i <= min(r, l + 2); i++) {
    a[i % 3] += (r - i) / 3 + 1;
  }
  dp[0][1] = a[0];
  dp[1][1] = a[1];
  dp[2][1] = a[2];
  for (int i = 2; i <= n; i++) {
    dp[0][i] += dp[0][i - 1] * a[0] + dp[1][i - 1] * a[2] + dp[2][i - 1] * a[1];
    dp[1][i] += dp[0][i - 1] * a[1] + dp[1][i - 1] * a[0] + dp[2][i - 1] * a[2];
    dp[2][i] += dp[0][i - 1] * a[2] + dp[1][i - 1] * a[1] + dp[2][i - 1] * a[0];
    for (int j = 0; j < 3; j++) dp[j][i] %= mod;
  }
  cout << dp[0][n] << endl;
  return 0;
}
