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 <bits/stdc++.h>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

typedef long long cat;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N;
	string S;
	cin >> N >> S;
	S.insert(begin(S), '1');
	S +='1';
	int inner1 =0, inner0 =0;
	for(int i =1; i <= 2*N; i++) if(S[i-1] == S[i]) {
		if(S[i-1] == '1') inner1++;
		else inner0++;
	}
	if(inner1%2 != 0 || inner0%2 != 0) {
		cout << "No\n";
		return 0;
	}

	vector< vector<int> > T; // topology
	for(int i =0; i <= 2*N; i++) if(i == 0 || (S[i-1] == '0' && S[i] == '1')) {
		vector<int> v(1, i);
		i++;
		while(i <= 2*N && S[i] == '1') {
			v.push_back(i);
			i++;
		}
		v.push_back(i);
		T.push_back(v);
	}
	vector<int> ans(2*N, -1);
	int col =1;

	if(inner1%4 != 0) {
		int seg =1;
		while(seg+1 < (int)T.size() && (int)T[seg].size() <= 2) seg++;
		if(seg+1 >= (int)T.size()) {
			cout << "No\n";
			return 0;
		}
		int s2 =-1;
		for(int i =0; i < (int)T.size(); i++) if((int)T[i].size() > 2 && i != seg) {
			s2 =i;
			break;
		}
		if(s2 == -1) {
			cout << "No\n";
			return 0;
		}
		ans[T[seg][0]-1] =ans[T[seg].back()-1] =col;
		col++;
		ans[T[s2][T[s2].size()-2]-1] =ans[T[seg][1]-1] =col;
		col++;
		int l =T[s2].back();
		T[s2].pop_back();
		T[s2].pop_back();
		for(int i =2; i < (int)T[seg].size()-1; i++) T[s2].push_back(T[seg][i]);
		T[s2].push_back(l);
		T.erase(begin(T)+seg);
	}

	// topology merge
	vector<int> Tfin;
	for(int i =0; i < (int)T.size(); i++) {
		if(i > 0) {
			ans[T[i-1].back()-1] =ans[T[i][0]-1] =col;
			col++;
		}
		for(int j =1; j < (int)T[i].size()-1; j++) Tfin.push_back(T[i][j]);
	}

	// 00 pairs
	int x0, p0sz =0;
	for(int i =1; i <= 2*N; i++) if(S[i] == '0' && S[i-1] == '0') {
		if(p0sz == 0) {
			x0 =i;
			p0sz++;
		}
		else {
			ans[x0-1] =ans[i-1] =col;
			p0sz =0;
			col++;
		}
	}

	// 11 pairs
	int x1[3];
	int p1sz =0;
	for(int i =0; i < (int)Tfin.size(); i++) {
		if(p1sz < 3) {
			x1[p1sz] =Tfin[i];
			p1sz++;
		}
		else {
			ans[x1[0]-1] =ans[x1[2]-1] =col;
			col++;
			ans[x1[1]-1] =ans[Tfin[i]-1] =col;
			col++;
			p1sz =0;
		}
	}

	cout << "Yes\n";
	for(int i =0; i < 2*N; i++) cout << ans[i] << ((i == 2*N-1)?"\n":" ");
	return 0;}

// look at my code
// my code is amazing