Problem: There is a bridge that connects the left and right banks of a river. There are 2 N doors placed at different positions on this bridge, painted in some colors. The colors of the doors are represented by integers from 1 through N. For each k (1 \leq k \leq N), there are exactly two doors painted in Color k.

Snuke decides to cross the bridge from the left bank to the right bank. He will keep on walking to the right, but the following event will happen while doing so:

* At the moment Snuke touches a door painted in Color k (1 \leq k \leq N), he teleports to the right side of the other door painted in Color k.



It can be shown that he will eventually get to the right bank.

For each i (1 \leq i \leq 2 N - 1), the section between the i-th and (i + 1)-th doors from the left will be referred to as Section i. After crossing the bridge, Snuke recorded whether or not he walked through Section i, for each i (1 \leq i \leq 2 N - 1). This record is given to you as a string s of length 2 N - 1. For each i (1 \leq i \leq 2 N - 1), if Snuke walked through Section i, the i-th character in s is `1`; otherwise, the i-th character is `0`.

<image>

Figure: A possible arrangement of doors for Sample Input 3

Determine if there exists an arrangement of doors that is consistent with the record. If it exists, construct one such arrangement.

Constraints

* 1 \leq N \leq 10^5
* |s| = 2 N - 1
* s consists of `0` and `1`.

Input

Input is given from Standard Input in the following format:


N
s


Output

If there is no arrangement of doors that is consistent with the record, print `No`. If there exists such an arrangement, print `Yes` in the first line, then print one such arrangement in the second line, in the following format:


c_1 c_2 ... c_{2 N}


Here, for each i (1 \leq i \leq 2 N), c_i is the color of the i-th door from the left.

Examples

Input

2
010


Output

Yes
1 1 2 2


Input

2
001


Output

No


Input

3
10110


Output

Yes
1 3 2 1 2 3


Input

3
10101


Output

No


Input

6
00111011100


Output

Yes
1 6 1 2 3 4 4 2 3 5 6 5

Solution (in CPP):

Solution:
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>

#include <cstdio>
#include <cstdlib>
#include <cmath>

#include <vector>
#include <deque>
#include <list>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <utility>

#include <string>
#include <cstring>

using namespace std;

#define LL long long
#define LD long double
#define SC(tp,var) static_cast < tp > ( var )
#define AR(tp) vector < tp >
#define PII pair < int, int >
#define PIL pair < int, long long >
#define PLI pair < long long, int >
#define PLL pair < long long, long long >
#define MP make_pair
#define FF first
#define SS second
#define PB push_back
#define PF push_front
#define POB pop_back
#define POF pop_front
#define PQ priority_queue
#define INIT(var,val) memset ( var, val, sizeof ( var ) )
#define fo(fname) freopen ( fname ".in", "r", stdin ); freopen ( fname ".out", "w", stdout );
#define lp(lpvar,lpst,lpen) for ( int lpvar = lpst, _lpen = lpen; lpvar < _lpen; ++lpvar )
#define lpi(lpvar,lpst,lpen) for ( int lpvar = lpst, _lpen = lpen; lpvar <= _lpen; ++lpvar )
#define lpd(lpvar,lpst,lpen) for ( int lpvar = lpst, _lpen = lpen; lpvar > _lpen; --lpvar )
#define lpdi(lpvar,lpst,lpen) for ( int lpvar = lpst, _lpen = lpen; lpvar >= _lpen; --lpvar )
#define qmax(a,b) (((a)<(b))?(b):(a))
#define qmin(a,b) (((a)<(b))?(a):(b))
#define qabs(a) (((a)<0)?(-(a)):(a))
#define qsqr(a) ((a)*(a))

const int INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffffLL;
const long long SLINF = 0x7fffffffffffffffLL;
const int MINF = 0x3f3f3f3f;
const LD DINF = pow ( 2, 100 );
const int MAXN = 400007;

int n, m;
char s[MAXN];
bool nd[MAXN];
int bfp[MAXN], kb;
int ans[MAXN], ka;
int gdp[MAXN], kg;
int p0[MAXN], kp0;



void init ();
void input ();
void work ();

void getmd ();



namespace task_bf
{
	const int MAXN = 100;
	int n;
	bool tar[MAXN];
	int cur[MAXN], p[MAXN];
	bool vis[MAXN];
	
	bool check ()
	{
		INIT ( vis, false );

		int now = 1;
		while ( now <= 2 * n ){
			now = p[now];
			vis[now] = true;
			++now;
		}
		lp ( i, 1, 2*n ) if ( vis[i] != tar[i] ) return false;
		return true;
	}

	bool srch ( int ni )
	{
		if ( ni > n ) return check ();

		lpi ( i, 1, n*2 ){
			if ( !cur[i] ){
				lpi ( j, i+1, n*2 ){
					if ( !cur[j] ){
						cur[i] = cur[j] = ni;
						p[i] = j, p[j] = i;
						if ( srch ( ni+1 ) ) return true;
						cur[i] = cur[j] = 0;
					}
				}
				break;
			}
		}

		return false;
	}

	bool work ()
	{
//		cerr << "task_bf::work : "; lpi ( i, 1, n ) cerr << tar[i]; cerr << endl;
		int tn = n;
		n = n >> 1;
		INIT ( cur, 0 );
		bool ans = srch ( 1 );
		n = tn;
		return ans;
	}
}

namespace task_gd
{
	int n;
	bool tar[MAXN];
	int ans[MAXN];
	int p1[MAXN], kp1;
	
	void work ()
	{
//		cerr << "task_gd : "; lpi ( i, 1, n ) cerr << tar[i]; cerr << endl;
		tar[0] = 1;
		int ka = 0;
		int c0 = -1;
		lpi ( i, 1, n ){
			if ( tar[i] ){
				if ( tar[i-1] ) p1[++kp1] = i;
				else{
					ans[c0] = ans[i] = ++ka;
					c0 = -1;
				}
			}else if ( c0 == -1 ) c0 = i;
		}
//		cerr << "p1 : "; lpi ( i, 1, kp1 ) cerr << p1[i] << " "; cerr << endl;
		for ( int i = 1; i <= kp1; i += 4 ) ans[p1[i]] = ans[p1[i+2]] = ++ka, ans[p1[i+1]] = ans[p1[i+3]] = ++ka;
	}
}



int main ()
{
	init ();
	input ();
	work ();
}



void init ()
{
	ios::sync_with_stdio ( false );
	cin.tie ( 0 );

//	fo ( "ya" );
}

void input ()
{
	scanf ( "%d%s", &n, s+1 );
}

void work ()
{
	m = n << 1;
	lp ( i, 1, m ) nd[i] = s[i] == '1';
	nd[0] = nd[m] = true;

	int c1 = 0;
	lpi ( i, 1, m ) if ( nd[i] && nd[i-1] ) ++c1;

	bool flag = true;
	bool full = false;
	if ( c1 & 1 ) flag = false;
	else if ( c1 >> 1 & 1 ){
		flag = false;
		int c0 = -1, nc = 1;
//		bool t0 = true;
		c1 = 0;
		lpi ( i, 1, m ){
			if ( nd[i] ){
				++nc;
				if ( nc >= 8 && nc % 4 == 0 ){
					full = true;
					task_bf::n -= 4;
					kb -= 4;
				}
//				if ( nc >= 2 ) t0 = true;
				if ( nc > 2 || ( nc == 2 && c0 == -1 ) ){
					task_bf::tar[++task_bf::n] = 1;
					bfp[++kb] = i;
					++c1;
				}else if ( nc == 2 ){
					task_bf::tar[++task_bf::n] = 0;
					task_bf::tar[++task_bf::n] = 1;
					bfp[++kb] = c0, bfp[++kb] = i-1;
					task_bf::tar[++task_bf::n] = 1;
					bfp[++kb] = i;
					++c1;
				}else if ( c1 % 4 == 2 ){
					task_bf::tar[++task_bf::n] = 0;
					task_bf::tar[++task_bf::n] = 1;
					bfp[++kb] = c0, bfp[++kb] = i;
					if ( task_bf::work () ){
						getmd ();
						flag = true;
						break;
					}
					task_bf::n -= 2;
					kb -= 2;
				}
				if ( !full && c1 % 4 == 2 ){
					if ( task_bf::work () ){
						getmd ();
						flag = true;
						break;
					}
				}
			}else if ( nc ) c0 = i, nc = 0, full = false;
/*				if ( nc ){
					c0 = i;
					nc = 0;
					if ( t0 && c1 % 4 == 2 ){
						task_bf::tar[++task_bf::n] = 0;
						bfp[++kb] = i;
						if ( task_bf::work () ){
							getmd ();
							flag = true;
							break;
						}
						--task_bf::n;
						--kb;
						t0 = false;
					}
				}*/
//			cerr << "after " << i << " : "; lpi ( i, 1, task_bf::n ) cerr << task_bf::tar[i]; cerr << endl;
		}
	}

	if ( !flag ){
		printf ( "No\n" );
		return;
	}
	
	lpi ( i, 1, m ){
		if ( !nd[i] && !nd[i-1] ){
			p0[++kp0] = i;
		}
	}
//	cerr << "p0 : "; lpi ( i, 1, kp0 ) cerr << p0[i] << " "; cerr << endl;
	for ( int i = 1; i <= kp0; i += 2 ) ans[p0[i]] = ans[p0[i+1]] = ++ka;

	lpi ( i, 1, m ){
		if ( !ans[i] ){
			task_gd::tar[++task_gd::n] = nd[i];
			gdp[++kg] = i;
		}
	}
	task_gd::work ();
	lpi ( i, 1, kg ){
		ans[gdp[i]] = task_gd::ans[i] + ka;
	}

	printf ( "Yes\n" );
	lpi ( i, 1, m ) printf ( "%d ", ans[i] );
	putchar ( '\n' );
}



void getmd ()
{
//	cerr << "getmd : " << endl;
//	lpi ( i, 1, kb ) cerr << bfp[i] << " "; cerr << endl;
//	lpi ( i, 1, kb ) cerr << task_bf::tar[i]; cerr << endl;
//	lpi ( i, 1, kb ) cerr << task_bf::cur[i] << " "; cerr << endl;
	lpi ( i, 1, kb ) ans[bfp[i]] = task_bf::cur[i];
	ka = kb >> 1;
}