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;
using ll = long long int;
template <typename T, typename U>
inline istream &operator>>(istream &in, pair<T, U> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T>
inline istream &operator>>(istream &in, vector<T> &v) {
  for (T &x : v) in >> x;
  return in;
}
const ll MOD = 1e9 + 7;
inline ll md(ll x, ll M = MOD) {
  x %= M;
  return ((x < 0) ? (x + M) : x);
}
ll f(ll l, ll r, ll rem, ll mod = 3) {
  ll ans = 0;
  ans += ((r + rem) / mod);
  if (l) ans -= ((l - 1 + rem) / mod);
  return ans;
}
void solve(ll &tc) {
  ll n, l, r;
  cin >> n >> l >> r;
  ll dp[n + 1][3];
  memset(dp, 0, sizeof(dp));
  for (ll j = 0; j < 3; j++) {
    dp[1][j] = f(l, r, j);
  }
  for (ll i = 2; i <= n; i++) {
    for (ll j = 0; j < 3; j++) {
      for (ll k = 0; k < 3; k++) {
        for (ll m = 0; m < 3; m++) {
          if ((k + m) % 3 == j) {
            dp[i][j] = md(dp[i][j] + dp[i - 1][k] * f(l, r, m));
          }
        }
      }
    }
  }
  cout << dp[n][0] << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll rep = 3;
  while (rep--) {
    ll tc = 1;
    for (ll i = 1; i <= tc; i++) {
      if (0) cerr << "Case #" << i << "\n";
      solve(i);
    }
    if (1) break;
  }
  return 0;
}
