Problem:
Score : 200 points

Problem Statement

Takahashi is competing in a sumo tournament.
The tournament lasts for 15 days, during which he performs in one match per day.
If he wins 8 or more matches, he can also participate in the next tournament.

The matches for the first k days have finished.
You are given the results of Takahashi's matches as a string S consisting of o and x.
If the i-th character in S is o, it means that Takahashi won the match on the i-th day; if that character is x, it means that Takahashi lost the match on the i-th day.

Print YES if there is a possibility that Takahashi can participate in the next tournament, and print NO if there is no such possibility.

Constraints

1 \leq k \leq 15

S is a string of length k consisting of o and x.

Input

Input is given from Standard Input in the following format:

S

Output

Print YES if there is a possibility that Takahashi can participate in the next tournament, and print NO otherwise.

Sample Input 1

oxoxoxoxoxoxox

Sample Output 1

YES

Takahashi has 7 wins and 7 losses before the last match. If he wins that match, he will have 8 wins.

Sample Input 2

xxxxxxxx

Sample Output 2

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize "-O3"
#pragma GCC optimize("Ofast")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define LSOne(S) (S & (-S))
#define ll long long
#define two pair<int,int>
#define twol pair<ll,ll>
#define four pair<two,two>
#define pb push_back
#define mk make_pair
#define y1 y1922
#define INF 1000000000000000000
#define P 1000000007
#define lmax 1000000000
#define nn 1000003
#define ff first.first
#define fs first.second
#define sf second.first
#define ss second.second
#define f first
#define s second
#define vi vector<int>
#define vll vector<ll>
#define vtwo vector<two>
#define ALL(container) (container).begin(), (container).end()
#define sz(container) (int)(container.size())
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define mid(a,b) (a+b>>1)
#define minN 0
#define maxN 10000000
#define na(x) ((x)<P?(x):(x)-P)
#define ab(a) (-(a)<(a)?(a):-(a))
#define FAST std::ios::sync_with_stdio(false)
#define xRand mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define rnd rng
#define IT iterator
typedef
tree<
  int,// aq pair<int,int> shegidzlia
  null_type,
  less/*_equal*/<int>,// aqac
  rb_tree_tag,
  tree_order_statistics_node_update>
ordered_set;
// '_equal' mashin ginda roca multiset gchirdeba
template<class key, class value,class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
ordered_map<int, int> my_map;
inline int rin(){
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
inline int bin(){
	int x=0;char ch=getchar();
	while (ch<'0'||ch>'9') ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return x;
}
string s;
int o;
int main(){FAST;xRand;
cin>>s;
for(int i=0;i<sz(s);i++){
	if(s[i]=='x') o++;
}
if(o<8){
	cout<<"YES"<<endl;
}else{
	cout<<"NO"<<endl;
}
}
/*

                   *         *
                  * *       * *
                 *   *     *   *
                *     *   *     *
                 *   *   * *   *
                  *   *   *   *
                   *   * *   *
                     *  *   *
					   *  *


*/