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 double eps = 1e-6;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long mod = 1e9 + 7;
vector<long long> kitne(long long l, long long r) {
  vector<long long> cnt(3, 0);
  if (r - l + 1 <= 10) {
    for (long long i = l; i <= r; ++i) {
      cnt[i % 3]++;
    }
    return cnt;
  }
  while (l % 3 != 0) cnt[l % 3]++, l++;
  while (r % 3 != 2) cnt[r % 3]++, --r;
  long long range = r - l + 1;
  range /= 3;
  cnt[0] += range;
  cnt[1] += range;
  cnt[2] += range;
  return cnt;
}
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  vector<long long> avail = kitne(l, r);
  long long dp[n + 4][3];
  memset(dp, 0, sizeof dp);
  for (long long i = 0; i < 3; ++i) dp[1][i] = avail[i];
  for (long long i = 2; i <= n; ++i) {
    dp[i][0] = avail[1] * dp[i - 1][2] + avail[2] * dp[i - 1][1] +
               avail[0] * dp[i - 1][0];
    dp[i][1] = avail[2] * dp[i - 1][2] + avail[1] * dp[i - 1][0] +
               avail[0] * dp[i - 1][1];
    dp[i][2] = avail[1] * dp[i - 1][1] + avail[0] * dp[i - 1][2] +
               avail[2] * dp[i - 1][0];
    dp[i][0] %= mod;
    dp[i][1] %= mod;
    dp[i][2] %= mod;
  }
  cout << dp[n][0] << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t = 1;
  for (long long i = 1; i <= t; ++i) {
    solve();
  }
}
