Problem:
Score : 2000 points

Problem Statement

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.

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.

Sample Input 1

2
010

Sample Output 1

Yes
1 1 2 2

Sample Input 2

2
001

Sample Output 2

No

Sample Input 3

3
10110

Sample Output 3

Yes
1 3 2 1 2 3

The figure below is identical to the one in the statement.

Sample Input 4

3
10101

Sample Output 4

No

Sample Input 5

6
00111011100

Sample Output 5

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

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

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld; 
typedef double db; 
typedef string str; 

typedef pair<int, int> pi;
typedef pair<ll,ll> pl; 
typedef pair<ld,ld> pd; 
#define mp make_pair
#define f first
#define s second

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<ld> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi; 
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back() 
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353; // = (119<<23)+1
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = 4*atan((ld)1); 
const int xd[4] = {0,1,0,-1}, yd[4] = {1,0,-1,0}; 

template<class T> bool ckmin(T& a, const T& b) { 
	return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; }
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());

#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

template <class T> using Tree = tree<T, null_type, less<T>, 
	rb_tree_tag, tree_order_statistics_node_update>; 
// change null_type for map
#define ook order_of_key
#define fbo find_by_order

void treeExample() {
	Tree<int> t, t2; t.insert(8);
	auto it = t.insert(10).f; assert(it == t.lb(9));
	assert(t.ook(10) == 1); assert(t.ook(11) == 2);
	assert(*t.fbo(0) == 8);
	t.join(t2); // assuming T < T2 or T > T2, merge t2 into t
}

namespace input {
	template<class T> void re(complex<T>& x);
	template<class T1, class T2> void re(pair<T1,T2>& p);
	template<class T> void re(vector<T>& a);
	template<class T, size_t SZ> void re(array<T,SZ>& a);

	template<class T> void re(T& x) { cin >> x; }
	void re(double& x) { string t; re(t); x = stod(t); }
	void re(ld& x) { string t; re(t); x = stold(t); }
	template<class T, class... Ts> void re(T& t, Ts&... ts) { 
		re(t); re(ts...); 
	}

	template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
	template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
	template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
	template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

using namespace input;

namespace output {
	void pr(int x) { cout << x; }
	void pr(long x) { cout << x; }
	void pr(ll x) { cout << x; }
	void pr(unsigned x) { cout << x; }
	void pr(unsigned long x) { cout << x; }
	void pr(unsigned long long x) { cout << x; }
	void pr(float x) { cout << x; }
	void pr(double x) { cout << x; }
	void pr(ld x) { cout << x; }
	void pr(char x) { cout << x; }
	void pr(const char* x) { cout << x; }
	void pr(const string& x) { cout << x; }
	void pr(bool x) { pr(x ? "true" : "false"); }
	template<class T> void pr(const complex<T>& x) { cout << x; }
	
	template<class T1, class T2> void pr(const pair<T1,T2>& x);
	template<class T> void pr(const T& x);
	
	template<class T, class... Ts> void pr(const T& t, const Ts&... ts) { 
		pr(t); pr(ts...); 
	}
	template<class T1, class T2> void pr(const pair<T1,T2>& x) { 
		pr("{",x.f,", ",x.s,"}"); 
	}
	template<class T> void pr(const T& x) { 
		pr("{"); // const iterator needed for vector<bool>
		bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; 
		pr("}");
	}
	
	void ps() { pr("\n"); } // print w/ spaces
	template<class T, class... Ts> void ps(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(" "); ps(ts...); 
	}
	
	void pc() { pr("]\n"); } // debug w/ commas
	template<class T, class... Ts> void pc(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(", "); pc(ts...); 
	}
	#define dbg(x...) pr("[",#x,"] = ["), pc(x);
}

using namespace output;

namespace io {
	void setIn(string s) { freopen(s.c_str(),"r",stdin); }
	void setOut(string s) { freopen(s.c_str(),"w",stdout); }
	void setIO(string s = "") {
		ios_base::sync_with_stdio(0); cin.tie(0); // fast I/O
		// cin.exceptions(cin.failbit); // ex. throws exception when you try to read letter into int
		if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
	}
}

using namespace io;

struct mi {
	typedef decay<decltype(MOD)>::type T;
	T val; 
	explicit operator T() const { return val; }
	mi() { val = 0; }
	mi(ll v) { 
		val = (-MOD <= v && v <= MOD) ? v : v % MOD;
		if (val < 0) val += MOD;
	}
	friend bool operator==(const mi& a, const mi& b) { 
		return a.val == b.val; }
	friend bool operator!=(const mi& a, const mi& b) { 
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { 
		return a.val < b.val; }
	friend void re(mi& a) { ll x; re(x); a = mi(x); }
	friend void pr(const mi& a) { pr(a.val); }
	friend ostream& operator<<(ostream& os, const mi& a) { 
		return os << a.val; }
   
	mi operator-() const { return mi(-val); }
	mi& operator+=(const mi& m) { 
		if ((val += m.val) >= MOD) val -= MOD; 
		return *this; }
	mi& operator-=(const mi& m) { 
		if ((val -= m.val) < 0) val += MOD; 
		return *this; }
	mi& operator++() { return *this += 1; }
	mi& operator--() { return *this -= 1; }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }

	mi& operator*=(const mi& m) { 
		val = (ll)val*m.val%MOD; return *this; }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) { 
		assert(!(a == 0)); return pow(a,MOD-2); }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};
typedef pair<mi,mi> pmi;
typedef vector<mi> vmi;
typedef vector<pmi> vpmi;

int N,cnt; str s;
vi c;

void assign(int a, int b) {
	assert(!c[a] && !c[b]);
	c[a] = c[b] = ++cnt;
}

void fin() {
	ps("Yes");
	vi bad; F0R(i,2*N) if (!c[i]) bad.pb(i);
	for (int i = 0; i < sz(bad); i += 2) assign(bad[i],bad[i+1]);
	F0R(i,2*N) {
		pr(c[i],' ');
	}
	exit(0);
}

void bad() {
	ps("No");
	exit(0);
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
	re(N,s); s += "1"; c.rsz(2*N);
	bool all = 1; trav(t,s) if (t == '0') all = 0;
	if (all) {
		if (N&1) bad();
		for (int i = 0; i < 2*N; i += 4) {
			assign(i,i+2);
			assign(i+1,i+3);
		}
		fin();
	}
	vector<vi> path;
	F0R(i,2*N) {
		int j = (i+2*N-1)%(2*N);
		if (s[i] == '1' && s[j] == '0') {
			int I = i; vi v;
			while (s[I] == '1') {
				v.pb(I);
				I = (I+1)%(2*N);
			}
			v.pb(I);
			path.pb(v);
		}
	}
	int len = 0;
	trav(t,path) len += sz(t)-2;
	if (len&1) bad();
	if (len%4 == 2) {
		vi cool;
		F0R(i,sz(path)) if (sz(path[i]) > 2) cool.pb(i);
		if (sz(cool) < 2) bad();
		assign(path[cool[0]][1],path[cool[1]][1]);
		swap(path[cool[0]][0],path[cool[1]][0]);
		path[cool[0]].erase(begin(path[cool[0]])+1);
		path[cool[1]].erase(begin(path[cool[1]])+1);
	}
	FOR(i,1,sz(path)) {
		assign(path[0].bk, path[i].ft);
		path[0].pop_back(), 
		path[0].insert(end(path[0]),begin(path[i])+1,end(path[i]));
	}
	assign(path[0].ft,path[0].bk);
	for (int i = 1; i < sz(path[0])-1; i += 4) {
		assign(path[0][i],path[0][i+2]);
		assign(path[0][i+1],path[0][i+3]);
	}
	fin();
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?), slow multiset operations
	* do smth instead of nothing and stay organized
*/