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>
const double pi = 3.14159265358979323846;
using namespace std;
long long fastexp(long long a, long long b) {
  long long answer = 1;
  long long temp = a;
  while (b > 0) {
    if (b & 1) answer *= (temp);
    temp *= temp;
    b >>= 1;
  }
  return answer;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  ;
  long long** dp = new long long*[3];
  for (long long i = 0; i < 3; i++) {
    dp[i] = new long long[n + 1];
  }
  dp[0][0] = 1;
  dp[1][0] = 1;
  dp[2][0] = 1;
  long long count_0, count_1, count_2;
  count_2 = 0;
  count_1 = 0;
  count_0 = 0;
  long long l;
  cin >> l;
  ;
  long long r;
  cin >> r;
  ;
  long long u_0, u_1, u_2;
  long long l_0, l_1, l_2;
  u_0 = u_1 = u_2 = (r / 3);
  r %= 3;
  ;
  if (r == 1)
    u_1++;
  else if (r == 2) {
    u_1++;
    u_2++;
  }
  l--;
  l_0 = l_2 = l_1 = (l / 3);
  l %= 3;
  if (l == 1)
    l_1++;
  else if (l == 2) {
    l_1++;
    l_2++;
  }
  count_0 = u_0 - l_0;
  count_1 = u_1 - l_1;
  count_2 = u_2 - l_2;
  dp[0][1] = count_0;
  dp[1][1] = count_1;
  dp[2][1] = count_2;
  for (long long i = 2; i <= n; i++) {
    long long j = 0;
    dp[0][i] = (dp[0][i - 1] * count_0 + dp[1][i - 1] * count_2 +
                dp[2][i - 1] * count_1) %
               1000000007;
    dp[1][i] = (dp[0][i - 1] * count_1 + dp[1][i - 1] * count_0 +
                dp[2][i - 1] * count_2) %
               1000000007;
    dp[2][i] = (dp[0][i - 1] * count_2 + dp[1][i - 1] * count_1 +
                dp[2][i - 1] * count_0) %
               1000000007;
  }
  cout << dp[0][n] << endl;
  return 0;
}
