Problem: You have a set of dominoes. Each domino is a rectangular tile with a line dividing its face into two square ends. Can you put all dominoes in a line one by one from left to right so that any two dominoes touched with the sides that had the same number of points? You can rotate the dominoes, changing the left and the right side (domino "1-4" turns into "4-1").

Input

The first line contains number n (1 ≤ n ≤ 100). Next n lines contains the dominoes. Each of these lines contains two numbers — the number of points (spots) on the left and the right half, correspondingly. The numbers of points (spots) are non-negative integers from 0 to 6.

Output

Print "No solution", if it is impossible to arrange the dominoes in the required manner. If the solution exists, then describe any way to arrange the dominoes. You put the dominoes from left to right. In each of n lines print the index of the domino to put in the corresponding position and then, after a space, character "+" (if you don't need to turn the domino) or "–" (if you need to turn it).

Examples

Input

5
1 2
2 4
2 4
6 4
2 1


Output

2 -
1 -
5 -
3 +
4 -

Solution (in CPP):

Solution:
//begin #include <Core>
/*
 * Package: StandardCodeLibrary.Core
 * */
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <utility>
#include <vector>
#include <list>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <numeric>
#include <bitset>
#include <complex>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <climits>
#if __GNUC__>=4 and __GNUC_MINOR__>=6
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#endif
using namespace std;

#define lp for(;;)
#define repf(i,a,b) for (int i=(a);i<(b);++i)
#define rep(i,n) repf(i,0,n)
#define ft(i,a,b) for (int i=(a);i<=(b);++i)
#define fdt(i,a,b) for (int i=(a);i>=b;--i)
#define feach(e,s) for (typeof((s).begin()) e=(s).begin();e!=(s).end();++e)
#define fsubset(subset,set) for (int subset=(set)&((set)-1);subset;subset=(subset-1)&(set))
#define forin(i,charset) for (cstr i=(charset);*i;i++)
#define whl while
#define rtn return
#define fl(x,y) memset((x),char(y),sizeof(x))
#define clr(x) fl(x,char(0))
#define cpy(x,y) memcpy(x,y,sizeof(x))
#define pb push_back
#define mp make_pair
#define ins insert
#define ers erase
#define lb lower_bound
#define ub upper_bound
#define rnk order_of_key
#define sel find_by_order
#define x first
#define y second
#define sz(x) (int((x).size()))
#define all(x) (x).begin(),(x).end()
#define srt(x) sort(all(x))
#define uniq(x) srt(x),(x).erase(unique(all(x)),(x).end())
#define vec vector
#define pr pair
#define que queue
#define prq priority_queue
#define itr iterator
#define sf scanf
#define pf printf
#define pdb(prcs,x) (cout<<setprecision(prcs)<<fixed<<(sgn(x)?(x):0))
#ifdef DEBUG
#define prt(x) cerr<<#x"="<<(x)<<endl
#define asrtWA(s) do if(!(s))do{cerr<<"assert("#s")"<<endl;}whl(0);whl(0)
#define asrtTLE(s) do if(!(s))do{cerr<<"assert("#s")"<<endl;}whl(0);whl(0)
#define asrtMLE(s) do if(!(s))do{cerr<<"assert("#s")"<<endl;}whl(0);whl(0)
#define asrtOLE(s) do if(!(s))do{cerr<<"assert("#s")"<<endl;}whl(0);whl(0)
#define asrtRE(s) do if(!(s))do{cerr<<"assert("#s")"<<endl;}whl(0);whl(0)
#define runtime() cerr<<"Used: "<<db(clock())/CLOCKS_PER_SEC<<"s"<<endl
#define input(in) do{}whl(0)
#define output(out) do{}whl(0)
#else
#define endl (char('\n'))
#define prt(x) (cerr)
#define asrtWA(s) do if(!(s))exit(0);whl(0)
#define asrtTLE(s) do if(!(s))whl(1);whl(0)
#define asrtMLE(s) do if(!(s))whl(new int);whl(0)
#define asrtOLE(s) do if(!(s))whl(1)puts("OLE");whl(0)
#define asrtRE(s) do if(!(s))*(int*)0=0;whl(0)
#define runtime() (cerr)
#define input(in) freopen(in,"r",stdin)
#define output(out) freopen(out,"w",stdout)
#endif

typedef long long int lli;
typedef double db;
typedef const char* cstr;
typedef string str;
typedef vec<int> vi;
typedef vec<vi> vvi;
typedef vec<bool> vb;
typedef vec<vb> vvb;
typedef vec<str> vs;
typedef pr<int,int> pii;
typedef pr<lli,lli> pll;
typedef pr<db,db> pdd;
typedef pr<str,int> psi;
typedef map<int,int> mii;
typedef map<str,int> msi;
typedef map<char,int> mci;
typedef set<int> si;
typedef set<str> ss;
typedef que<int> qi;
typedef prq<int> pqi;
#if __GNUC__>=4 and __GNUC_MINOR__>=7
template<typename key,typename value>class ext_map:public __gnu_pbds::tree<key,value,less<key>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>{};
template<typename key>class ext_set:public __gnu_pbds::tree<key,__gnu_pbds::null_type,less<key>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>{};
#elif __GNUC__>=4 and __GNUC_MINOR__>=6
template<typename key,typename value>class ext_map:public __gnu_pbds::tree<key,value,less<key>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>{};
template<typename key>class ext_set:public __gnu_pbds::tree<key,__gnu_pbds::null_mapped_type,less<key>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>{};
#endif

const int oo=(~0u)>>1;
const lli ooll=(~0ull)>>1;
const db inf=1e+10;
const db eps=1e-10;
const db pi=acos(-1.0);
const int MOD=1000000007;

template<typename type>inline bool cmax(type& a,const type& b){rtn a<b?a=b,true:false;}
template<typename type>inline bool cmin(type& a,const type& b){rtn b<a?a=b,true:false;}
template<typename type>inline type sqr(const type& x){rtn x*x;}
inline int dbcmp(const db& a,const db& b){rtn (a>b+eps)-(a<b-eps);}
inline int sgn(const db& x){rtn dbcmp(x,0);}
template<typename ostream,typename type>ostream& operator<<(ostream& cout,const pr<type,type>& x){rtn cout<<"("<<x.x<<","<<x.y<<")";}
template<typename type>pr<type,type> operator-(const pr<type,type>& x){rtn mp(-x.x,-x.y);}
template<typename type>pr<type,type> operator+(const pr<type,type>& a,const pr<type,type>& b){rtn mp(a.x+b.x,a.y+b.y);}
template<typename type>pr<type,type> operator-(const pr<type,type>& a,const pr<type,type>& b){rtn mp(a.x-b.x,a.y-b.y);}
template<typename type>inline type cross(const pr<type,type>& a,const pr<type,type>& b,const pr<type,type>& c){rtn (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}
template<typename type>inline type dot(const pr<type,type>& a,const pr<type,type>& b,const pr<type,type>& c){rtn (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);}
template<typename type>inline type gcd(type a,type b){if(b)whl((a%=b)&&(b%=a));rtn a+b;}
template<typename type>inline type lcm(type a,type b){rtn a*b/gcd(a,b);}
template<typename type>inline void bit_inc(vec<type>& st,int x,type inc){whl(x<sz(st))st[x]+=inc,x|=x+1;}
template<typename type>inline type bit_sum(const vec<type>& st,int x){type s=0;whl(x>=0)s+=st[x],x=(x&(x+1))-1;rtn s;}
template<typename type>inline type bit_kth(const vec<type>& st,int k){int x=0,y=0,z=0;whl((1<<(++y))<=sz(st));fdt(i,y-1,0){if((x+=1<<i)>sz(st)||z+st[x-1]>k)x-=1<<i;else z+=st[x-1];}rtn x;}
inline void make_set(vi& st){rep(i,sz(st))st[i]=i;}
inline int find_set(vi& st,int x){int y=x,z;whl(y!=st[y])y=st[y];whl(x!=st[x])z=st[x],st[x]=y,x=z;rtn y;}
inline bool union_set(vi& st,int a,int b){a=find_set(st,a),b=find_set(st,b);rtn a!=b?st[a]=b,true:false;}
template<typename type>inline void merge(type& a,type& b){if(sz(a)<sz(b))swap(a,b);whl(sz(b))a.insert(*b.begin()),b.erase(b.begin());}
template<typename type>inline void merge(prq<type>& a,prq<type>& b){if(sz(a)<sz(b))swap(a,b);whl(sz(b))a.push(b.top()),b.pop();}

struct Initializer{Initializer(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}~Initializer(){runtime();}}initializer;
//end #include <Core>

void go(vvi& g,vi& lst,int u)
{
	prt(u);
	rep(v,7) if (g[u][v])
	{
		--g[u][v],--g[v][u];
		go(g,lst,v);
	}
	lst.pb(u);
}

int main()
{
	int n;
	cin>>n;
	vvi g(7,vi(7));
	vec<vvi > st(7,vvi(7));
	rep(i,n)
	{
		int u,v;
		cin>>u>>v;
		++g[u][v],++g[v][u];
		st[u][v].pb(i+1);
	}
	int s=-1;
	vi cnt(2);
	rep(i,7)
	{
		int tol=0;
		rep(j,7) tol+=g[i][j];
		cnt[tol%2]++;
		if (tol%2==1) s=i;
		if (s==-1&&tol) s=i;
	}
	if (cnt[1]!=2&&cnt[1]!=0) cout<<"No solution"<<endl;
	else
	{
		vi lst;
		go(g,lst,s);
		if (sz(lst)!=n+1) cout<<"No solution"<<endl;
		else
			repf(i,1,sz(lst))
			{
				if (sz(st[lst[i-1]][lst[i]]))
				{
					cout<<st[lst[i-1]][lst[i]].back()<<" +"<<endl;
					st[lst[i-1]][lst[i]].pop_back();
				}
				else
				{
					cout<<st[lst[i]][lst[i-1]].back()<<" -"<<endl;
				st[lst[i]][lst[i-1]].pop_back();
				}
			}
	}
}