Alex doesn’t like boredom. That’s why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let’s denote it a k) and delete it, at that all elements equal to a k + 1 and a k - 1 also must be deleted from the sequence. That step brings a k points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input
The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex’s sequence.
The second line contains n integers a 1, a 2, …, a n (1 ≤ a i ≤ 105).
Output
Print a single integer — the maximum number of points that Alex can earn.
Examples
input
2
1 2
output
2
input
3
1 2 3
output
4
input
9
1 2 1 3 2 2 2 2 3
output
10
Note
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
Solution:
#include <cstring> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <memory.h> #include <cassert> using namespace std; const int N = 1000010; int cnt[N]; long long f[N]; int main() { int n; scanf("%d", &n); memset(cnt, 0, sizeof(cnt)); for (int i = 0; i < n; i++) { int foo; scanf("%d", &foo); cnt[foo]++; } f[0] = 0; for (int i = 1; i < N; i++) { f[i] = (long long)i * cnt[i]; if (i - 2 >= 0) { f[i] += f[i - 2]; } if (f[i - 1] > f[i]) { f[i] = f[i - 1]; } } cout << f[N - 1] << endl; return 0; }