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;
mt19937 rnd(1337);
ostream& operator<<(ostream& a, const pair<long long, long long>& b) {
  cout << "{" << b.first << ", " << b.second << "}";
  return a;
}
ostream& operator<<(ostream& a, const vector<long long int>& b) {
  for (auto& k : b) cout << k << " ";
  return a;
}
ostream& operator<<(ostream& a, const vector<pair<long long, long long> >& b) {
  for (auto& k : b) cout << k << " ";
  return a;
}
const long long int INF = (long long int)1e18;
const long long int MOD = 1000 * 1000 * 1000 + 7, MOD2 = 274876858367;
const long long int maxn = (long long int)2e5 + 10, L = 10;
long long int dp[maxn][3], cnt[3];
void add(long long int& a, long long int b) { a = (a + b) % MOD; }
long long int mul(long long int a, long long int b) { return (a * b) % MOD; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, l, r;
  cin >> n >> l >> r;
  for (int i = 0; i < 3; ++i) {
    cnt[i] = r >= i ? ((r - i) / 3 - (l - i + 2) / 3 + 1) : 0;
    dp[0][i] = cnt[i];
  }
  for (long long int i = 0; i < n - 1; ++i) {
    for (long long int j = 0; j < 3; ++j) {
      for (long long int k = 0; k < 3; ++k) {
        add(dp[i + 1][(j + k) % 3], mul(dp[i][j], cnt[k]));
      }
    }
  }
  cout << dp[n - 1][0];
  return 0;
}
