Row

time limit            per test 

1 second                memory limit                  per test256 megabytes

input                  standard input                     output                            standard output

You're given a row with $\mathrm{nn chairs.\; We\; call\; a\; seating\; of\; people\; "maximal"\; if\; the\; two\; following\; conditions\; hold:}$

1. There are no neighbors adjacent to anyone seated.
2. It's impossible to seat one more person without violating the first rule.

The seating is given as a string consisting of zeros and ones ($\mathrm{00 means\; that\; the\; corresponding\; seat\; is\; empty, 11 —\; occupied).\; The\; goal\; is\; to\; determine\; whether\; this\; seating\; is\; "maximal".}$

Note that the first and last seats are not adjacent (if $\mathrm{n\ne 2n\ne 2).}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 10001\le n\le 1000) —\; the\; number\; of\; chairs.}$

The next line contains a string of $\mathrm{nn characters,\; each\; of\; them\; is\; either\; zero\; or\; one,\; describing\; the\; seating.}$

Output

Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".

You are allowed to print letters in whatever case you'd like (uppercase or lowercase).

Examples

input

3
101

output

Yes

input

4
1011

output

No

input

5
10001

output

No

Note

In sample case one the given seating is maximal.

In sample case two the person at chair three has a neighbour to the right.

In sample case three it is possible to seat yet another person into chair three.

 

大意：0代表该座位没有人占，1代表该座位有人占。

　　　　有两个规则： 1，两个1之间必须隔一个0。

　　　　　　　　　　2，使1尽可能多。

特判一下前两个元素和后两个元素（wa了两发，没考虑后两个元素），中间的元素只需考虑 两个1之间只能隔1~2个0就行了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 100;
int T,n,m;

using namespace std;
int main()
{
    int i,n,d1=INF,flag=0;
    char s[1010]={0};
    char emp;
    scanf("%d%c",&n,&emp);
    gets(s);
    for(i=0;i<n;i++)
    {
        if(s[i]=='1')
        {
            d1=i;
            break;
        }
    }
    if(d1>1)
        flag=1;
    for(i=d1+1;i<n;i++)
    {
        if(s[i]=='1')
        {
            if( !((i-d1)==2||(i-d1)==3) ){
                flag=1;
                break;
            }
            else
                d1=i;
        }
    }
    if(n>=2)
    {
        if(s[n-1]=='0'&&s[n-2]=='0')
            flag=1;
    }
    if(flag)
        printf("No\n");
    else
        printf("Yes\n");
    return 0;
}