歐拉環 測資有點 tricky~
只要奇點數量不是 0 跟 2 就是無解
/\*
\* GCA : Where is the Dp,there is the wall
\*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#endif
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Pln() printf("\\n")
#define M 106
#define PB push\_back
#define oo INT\_MAX
#define X first
#define Y second
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-9
inline bool xdy(double x,double y){return x>y+eps;}
inline bool xddy(double x,double y){return x>y-eps;}
inline bool xcy(double x,double y){return x<y-eps;}
inline bool xcdy(double x,double y){return x<y+eps;}
int m;
int mz\[10\]\[10\];
int ind\[10\];
#define PIC pair<int,char\>
struct edges{
int x,y;
int vis;
edges(){
x=y=-1;
vis=0;
}
}edge\[M\];
vector<PIC> ans;
int pEdge(int x,int y){
for(int i=0;i<m;i++){
if(edge\[i\].x==x&&edge\[i\].y==y&&edge\[i\].vis==0){
edge\[i\].vis=1;
return i;
}else if(edge\[i\].x==y&&edge\[i\].y==x&&edge\[i\].vis==0){
edge\[i\].vis=1;
return i+m;
}
}
return -1;
}
void dfs(int now){
for(int i=0;i<=6;i++){
if(mz\[now\]\[i\]){
mz\[now\]\[i\]--;
mz\[i\]\[now\]--;
dfs(i);
int tmp=pEdge(now,i);
if(tmp>=m){
ans.PB(PIC(tmp-m,'-'));
}else{
ans.PB(PIC(tmp,'+'));
}
}
}
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d",&m)){
Set(mz,0);
Set(ind,0);
ans.clear();
for(int i=0;i<m;i++){
int x,y;
scanf("%d%d",&x,&y);
edge\[i\].vis=0;
edge\[i\].x=x;edge\[i\].y=y;
mz\[x\]\[y\]++,mz\[y\]\[x\]++;
ind\[x\]++,ind\[y\]++;
}
int fir=-1;
int cntodd=0;
for(int i=0;i<=6;i++){
if(ind\[i\]&1){
cntodd++;
fir=i;
}
}
if(fir==-1){
for(int i=0;i<=6;i++){
if(ind\[i\])fir=i;
}
}
if(cntodd!=0&&cntodd!=2){
puts("No solution");
continue;
}
dfs(fir);
if(ans.size()>=m)
for(int i=ans.size()-1;i>=0;i--){
int x=ans\[i\].X;
char y=ans\[i\].Y;
printf("%d %c\\n",x+1,y);
}
else puts("No solution");
}
}