Problem:
Score: 300 points

Problem Statement

In Republic of Atcoder, there are N prefectures, and a total of M cities that belong to those prefectures.

City i is established in year Y_i and belongs to Prefecture P_i.

You can assume that there are no multiple cities that are established in the same year.

It is decided to allocate a 12-digit ID number to each city.

If City i is the x-th established city among the cities that belong to Prefecture i, the first six digits of the ID number of City i is P_i, and the last six digits of the ID number is x.

Here, if P_i or x (or both) has less than six digits, zeros are added to the left until it has six digits.

Find the ID numbers for all the cities.

Note that there can be a prefecture with no cities.

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq P_i \leq N

1 \leq Y_i \leq 10^9

Y_i are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
P_1 Y_1
:
P_M Y_M

Output

Print the ID numbers for all the cities, in ascending order of indices (City 1, City 2, ...).

Sample Input 1

2 3
1 32
2 63
1 12

Sample Output 1

000001000002
000002000001
000001000001

As City 1 is the second established city among the cities that belong to Prefecture 1, its ID number is 000001000002.

As City 2 is the first established city among the cities that belong to Prefecture 2, its ID number is 000002000001.

As City 3 is the first established city among the cities that belong to Prefecture 1, its ID number is 000001000001.

Sample Input 2

2 3
2 55
2 77
2 99

Sample Output 2

000002000001
000002000002
000002000003

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

#include <bits/stdc++.h>

#define INF 100100100
#define MOD 1000000007

using namespace std;
using ll = long long;
using ull = unsigned long long ;

int main(int argc, char const* argv[]){
   ios::sync_with_stdio(false);
   cin.tie(0);

   ll N,M;
   cin>>N>>M;
   vector<vector<pair<int,int>> > si(N+1);
   ll MAX = -1;
   for(ll i=0;i<M;i++){
      ll P;
      ll Y;
      cin>>P>>Y;
      si[P].push_back(make_pair(Y,i));
   }
   vector<pair<ll,ll> > ans(M);
   for(int i=1;i<N+1;i++)sort(si[i].begin(),si[i].end());
   for(int i=1;i<N+1;i++){
      for(int j=0;j<si[i].size();j++){
         ans[si[i][j].second] = make_pair(i,j+1);
      }
   }
   for(int i=0;i<ans.size();i++)cout<<setfill('0')<<right<<setw(6)<<ans[i].first<<right<<setw(6)<<ans[i].second<<endl;
   return 0;
}