E. Boxers（贪心）

E. Boxers（贪心）

There are nn boxers, the weight of the ii-th boxer is aiai. Each of them can change the weight by no more than 11 before the competition (the weight cannot become equal to zero, that is, it must remain positive). Weight is always an integer number.

It is necessary to choose the largest boxing team in terms of the number of people, that all the boxers' weights in the team are different (i.e. unique).

Write a program that for given current values aiai will find the maximum possible number of boxers in a team.

It is possible that after some change the weight of some boxer is 150001150001 (but no more).

Input

The first line contains an integer nn (1≤n≤1500001≤n≤150000) — the number of boxers. The next line contains nn integers a1,a2,…,ana1,a2,…,an, where aiai (1≤ai≤1500001≤ai≤150000) is the weight of the ii-th boxer.

Output

Print a single integer — the maximum possible number of people in a team.

Examples

input

4
3 2 4 1

output

input

6
1 1 1 4 4 4

output

Note

In the first example, boxers should not change their weights — you can just make a team out of all of them.

In the second example, one boxer with a weight of 11 can be increased by one (get the weight of 22), one boxer with a weight of 44 can be reduced by one, and the other can be increased by one (resulting the boxers with a weight of 33 and 55, respectively). Thus, you can get a team consisting of boxers with weights of 5,4,3,2,15,4,3,2,1.

题意：有n个拳击运动员，每个人在比赛前可以改变一斤体重，询问比赛时最多能带几个不同体重的运动员。

思路：先排序，从大到小，优先想加一斤体重，再不变，最后考虑减一斤体重。

代码：

#include <bits/stdc++.h>using namespace std;typedef long long ll;
const int N = 2e5+10;int via[N];
int a[N];
int n;int main()
{int i,j,k;scanf("%d", &n);for ( i=1; i<=n; i++ ) {cin >> a[i];}sort(a+1,a+n+1,greater<int>());int ans = 0;for ( i=1; i<=n; i++ ) {if ( via[a[i]+1]==0 ) {via[a[i]+1] = 1;ans ++;}else if ( via[ a[i] ]==0 ) {via[ a[i] ] = 1;ans ++;}else if ( via[ a[i]-1 ]==0 && a[i]-1>0 ) {via[ a[i]-1 ] = 1;ans ++;}}cout << ans << endl;return 0;
}

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