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 = (long long)1e9 + 7;
int n, l, r;
long long dp[200009][3];
int ck(int t, int v) {
  int retval = 0;
  retval = t / 3;
  if (v == 1) {
    if (t % 3 >= 1) retval++;
  } else if (v == 2) {
    if (t % 3 >= 2) retval++;
  }
  return retval;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> r;
  dp[1][0] = ck(r, 0) - ck(l - 1, 0);
  dp[1][1] = ck(r, 1) - ck(l - 1, 1);
  dp[1][2] = ck(r, 2) - ck(l - 1, 2);
  for (int i = 2; i <= n; i++) {
    dp[i][0] = dp[i - 1][0] * dp[1][0];
    dp[i][0] = dp[i][0] % MOD;
    dp[i][0] += dp[i - 1][1] * dp[1][2];
    dp[i][0] = dp[i][0] % MOD;
    dp[i][0] += dp[i - 1][2] * dp[1][1];
    dp[i][0] = dp[i][0] % MOD;
    dp[i][1] = dp[i - 1][0] * dp[1][1];
    dp[i][1] = dp[i][1] % MOD;
    dp[i][1] += dp[i - 1][1] * dp[1][0];
    dp[i][1] = dp[i][1] % MOD;
    dp[i][1] += dp[i - 1][2] * dp[1][2];
    dp[i][1] = dp[i][1] % MOD;
    dp[i][2] = dp[i - 1][0] * dp[1][2];
    dp[i][2] = dp[i][2] % MOD;
    dp[i][2] += dp[i - 1][1] * dp[1][1];
    dp[i][2] = dp[i][2] % MOD;
    dp[i][2] += dp[i - 1][2] * dp[1][0];
    dp[i][2] = dp[i][2] % MOD;
  }
  cout << dp[n][0] << "\n";
  return 0;
}
