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;
  vector<vector<long long> > dp(n + 1);
  for (int i = 0; i <= n; ++i) {
    dp[i].resize(3, 0);
  }
  long long tr = r / 3, tl = l / 3, tel = -1, ter;
  for (long long i = tl; i <= tr; i++) {
    if (3 * i >= l && 3 * i <= r) {
      tel = i;
      break;
    }
  }
  for (long long i = tr; i >= tl; i--) {
    if (3 * i >= l && 3 * i <= r) {
      ter = i;
      break;
    }
  }
  if (tel != -1) dp[1][0] = (ter - tel + 1) % 1000000007;
  tr = (r - 1) / 3;
  tl = (l - 1) / 3;
  tel = -1;
  for (long long i = tl; i <= tr; i++) {
    if (3 * i + 1 >= l && 3 * i + 1 <= r) {
      tel = i;
      break;
    }
  }
  for (long long i = tr; i >= tl; i--) {
    if (3 * i + 1 >= l && 3 * i + 1 <= r) {
      ter = i;
      break;
    }
  }
  if (tel != -1) dp[1][1] = (ter - tel + 1) % 1000000007;
  tr = (r - 2) / 3;
  tl = (l - 2) / 3;
  tel = -1;
  for (int i = tl; i <= tr; i++) {
    if (3 * i + 2 >= l && 3 * i + 2 <= r) {
      tel = i;
      break;
    }
  }
  for (int i = tr; i >= tl; i--) {
    if (3 * i + 2 >= l && 3 * i + 2 <= r) {
      ter = i;
      break;
    }
  }
  if (tel != -1) dp[1][2] = (ter - tel + 1) % 1000000007;
  for (int i = 2; i <= n; i++) {
    dp[i][0] = (dp[i - 1][0] * dp[1][0]) % 1000000007;
    dp[i][0] += (dp[i - 1][1] * dp[1][2]) % 1000000007;
    dp[i][0] += (dp[1][1] * dp[i - 1][2]) % 1000000007;
    dp[i][1] = (dp[i - 1][2] * dp[1][2]) % 1000000007;
    dp[i][1] += (dp[i - 1][1] * dp[1][0]) % 1000000007;
    dp[i][1] += (dp[1][1] * dp[i - 1][0]) % 1000000007;
    dp[i][2] = (dp[i - 1][1] * dp[1][1]) % 1000000007;
    dp[i][2] += (dp[i - 1][0] * dp[1][2]) % 1000000007;
    dp[i][2] += (dp[1][0] * dp[i - 1][2]) % 1000000007;
  }
  cout << dp[n][0] % 1000000007;
  return 0;
}
