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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const int INF = 100000000;

ll i,j,k;
int main(){
    ll n,m;cin>>n>>m;
    vector<ll> p(m),y(m);//p:所属する県 y:市の誕生年
    map<pair<ll,ll>,ll> answer;//pair:県と誕生年　ll:同じ県の中での誕生番号

    vector<pair<ll,ll>> yp_set(m),yp_set2(m);
    for(i=0;i<m;i++){
        cin>>p[i]>>y[i];
        yp_set[i]=make_pair(y[i],p[i]);//yとpのset
        yp_set2[i]=make_pair(y[i],p[i]);//yとpのset
        answer[yp_set2[i]]=0;
    }

//////県番号の小さい順　かつ　誕生番号の小さい順に並べる/////////////////////////////////////

    sort(yp_set.begin(),yp_set.end());//y(誕生年)でsort
    
    ll ins;
    for(i=0;i<m;i++){//y p swap
        ins=yp_set[i].first;
        yp_set[i].first=yp_set[i].second;
        yp_set[i].second=ins;
    }
    sort(yp_set.begin(),yp_set.end());

    for(i=0;i<m;i++){//y p swap(元に戻した)　　これで県と誕生番号が適切にsortされた
        ins=yp_set[i].first;
        yp_set[i].first=yp_set[i].second;
        yp_set[i].second=ins;
    }

    
/////////////////////////////////////////////////////////////////////////////////

    for(i=0;i<m-1;i++){
        if(i==0){answer[yp_set[i]]=1;}
        
        if(yp_set[i].second!=yp_set[i+1].second){answer[yp_set[i+1]]=1;}//違う県になったら、はじめは1にする
        else{answer[yp_set[i+1]]=answer[yp_set[i]]+1;}
    }//県内で何番目に誕生したか


    for(i=0;i<m;i++){
        answer[yp_set[i]]+=1000000*yp_set[i].second;
    }//県番号

    for(i=0;i<m;i++){
        printf("%012lld\n",answer[yp_set2[i]]);
    }
    return 0;
}