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 main() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long base = (r - l + 1) / 3;
  long long mod = 1000000007;
  long long freq[3];
  freq[0] = base;
  freq[1] = base;
  freq[2] = base;
  long long rem = (r - l + 1) % 3;
  for (long long i = 0; i < rem; ++i) {
    freq[(r - i) % 3]++;
  }
  if (r - l + 1 == 1) {
    if (r % 3 == 0 || n % 3 == 0)
      cout << 1;
    else
      cout << 0;
    return 0;
  }
  long long dp[n + 1][3];
  dp[0][0] = 1;
  dp[0][1] = 0;
  dp[0][2] = 0;
  for (long long i = 1; i <= n; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
    dp[i][2] = 0;
    for (long long j = 0; j < 3; j++) {
      if (freq[0]) {
        dp[i][j] += (dp[i - 1][j] * freq[0]) % mod;
        dp[i][j] %= mod;
      }
      dp[i][j] += (dp[i - 1][(j - 1 + 3) % 3] * freq[1]) % mod;
      dp[i][j] %= mod;
      dp[i][j] += (dp[i - 1][(j - 2 + 3) % 3] * freq[2]) % mod;
      dp[i][j] %= mod;
    }
  }
  cout << dp[n][0];
}
