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;
map<long long, long long> mp;
long long dp[200005][4], a[5];
long long n, l, r;
long long fun(long long pos, long long sum) {
  if (pos > n) {
    if (sum == 0) {
      return 1;
    }
    return 0;
  }
  if (dp[pos][sum] != -1) {
    return dp[pos][sum];
  }
  long long f1 = (mp[0] * (fun(pos + 1, sum % 3) % (long long)1000000007)) %
                 (long long)1000000007;
  long long f2 =
      (mp[1] * (fun(pos + 1, (sum + 1) % 3) % (long long)1000000007)) %
      (long long)1000000007;
  long long f3 =
      (mp[2] * (fun(pos + 1, (sum + 2) % 3) % (long long)1000000007)) %
      (long long)1000000007;
  long long ans = max(f1, 0LL);
  ans += max(f2, 0LL);
  ans += max(f3, 0LL);
  ans %= (long long)1000000007;
  return dp[pos][sum] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  while (cin >> n >> l >> r) {
    long long lo = l, hi = r, x;
    memset(dp, -1, sizeof(dp));
    while (lo <= r && lo % 3 != 0) {
      long long x = 3 - (lo % 3);
      mp[x]++;
      lo++;
    }
    while (hi > lo && hi % 3 != 0) {
      long long x = 3 - (hi % 3);
      mp[x]++;
      hi--;
    }
    x = (hi / 3) - ((lo - 1) / 3);
    x = max(0LL, x);
    mp[0] = x;
    mp[1] += max(x - 1, 0LL);
    mp[2] += max(x - 1, 0LL);
    cout << fun(1, 0) << endl;
    mp.clear();
  }
  return 0;
}
