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 = 1e9 + 7;
long long quick(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long cheng(long long a, long long b) {
  long long ans = 0;
  while (b) {
    if (b & 1) {
      ans = (ans + a) % mod;
    }
    a = (a + a) % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  long long n, l, r;
  cin >> n >> l >> r;
  if (l == r) {
    cout << ((l * n) % 3 == 0) << endl;
    return 0;
  }
  if (r - l == 1) {
    if (n == 1) {
      cout << (l % 3 != 1) << endl;
      return 0;
    }
    long long ans;
    if (l % 3 == 1) {
      ans = 2;
      int jia[2] = {2, -2};
      for (int i = 3; i <= n; i++) {
        ans = (ans * 2 + jia[i % 2]) % mod;
      }
    } else {
      ans = 1;
      int jia[6] = {0, -1, -1, 0, 1, 1};
      for (int i = 3; i <= n; i++) {
        ans = (ans * 2 + jia[i % 6]) % mod;
      }
    }
    cout << ans << endl;
    return 0;
  }
  if (l % 3 == (r + 1) % 3) {
    long long x = quick(r - l + 1, n - 1);
    long long a = (r - l + 1) / 3;
    a %= mod;
    cout << (x * a) % mod << endl;
    return 0;
  }
  if (l % 3 == 0 && r % 3 == 0) {
    long long x = quick(r - l + 1, n - 1);
    long long a = (r - l + 1) / 3;
    cout << (((cheng(3 * a, x) + x + 2) % mod) * 333333336) % mod << endl;
    return 0;
  }
  if (l % 3 != 0 && l % 3 == r % 3) {
    long long x = quick(r - l + 1, n - 1);
    long long a = (r - l + 1) / 3;
    if ((n - 1) % 3 != 2)
      cout << (((cheng(3 * a, x) + x - 1) % mod) * 333333336) % mod << endl;
    else
      cout << (((cheng(3 * a, x) + x + 2) % mod) * 333333336) % mod << endl;
  }
  if (l % 3 == 1 && r % 3 == 2) {
    long long x = quick(r - l + 1, n - 1);
    long long a = (r - l + 1) / 3;
    if ((n - 1) % 2) {
      cout << (((cheng(3 * a, x) + cheng(2, x) + 2) % mod) * 333333336) % mod
           << endl;
    } else {
      cout << (((cheng(3 * a, x) + cheng(2, x) - 2) % mod) * 333333336) % mod
           << endl;
    }
  }
  if (l % 3 != 1 && l % 3 == (r - 1) % 3) {
    long long x = quick(r - l + 1, n - 1);
    long long a = ((r - l) / 3) % mod;
    if ((n - 1) % 6 == 1 || (n - 1) % 6 == 3) {
      cout << (((cheng(3 * a, x) + cheng(2, x) - 1) % mod) * 333333336) % mod
           << endl;
    } else if ((n - 1) % 6 == 2) {
      cout << (((cheng(3 * a, x) + cheng(2, x) - 2) % mod) * 333333336) % mod
           << endl;
    } else if ((n - 1) % 6 == 4 || (n - 1) % 6 == 0) {
      cout << (((cheng(3 * a, x) + cheng(2, x) + 1) % mod) * 333333336) % mod
           << endl;
    } else if ((n - 1) % 6 == 5) {
      cout << (((cheng(3 * a, x) + cheng(2, x) + 2) % mod) * 333333336) % mod
           << endl;
    }
  }
  return 0;
}
