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;
long long int mod = 1e9 + 7;
long long int mul(long long int x, long long int y) {
  return (x * 1ll * y) % mod;
}
long long int add(long long int x, long long int y) {
  x += y;
  while (x >= mod) x -= mod;
  while (x < 0) x += mod;
  return x;
}
long long int dp[200005][3];
long long int a, b, c;
long long int n, l, r;
long long int fun(int i, int j) {
  if (i == n) {
    if (j == 0)
      return 1;
    else
      return 0;
  }
  if (dp[i][j] != -1) return dp[i][j];
  dp[i][j] = add(
      mul(a, fun(i + 1, (j) % 3)),
      add(mul(b, fun(i + 1, (j + 1) % 3)), mul(c, fun(i + 1, (j + 2) % 3))));
  return dp[i][j];
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> l >> r;
  long long int x = (l - 1 + 2) / 3;
  long long int y = (r - 1) / 3;
  b = y - x + 1;
  x = (l - 2 + 2) / 3;
  y = (r - 2) / 3;
  c = y - x + 1;
  x = (l + 2) / 3;
  y = (r) / 3;
  a = y - x + 1;
  if (l == 1 && r == 1) {
    a = 0, b = 1, c = 0;
  }
  memset(dp, -1, sizeof(dp));
  long long int ans = 0;
  ans = add(ans, mul(a, fun(1, 0)));
  ans = add(ans, mul(b, fun(1, 1)));
  ans = add(ans, mul(c, fun(1, 2)));
  cout << ans << endl;
}
