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>
typedef long long ll;
using namespace std;
ll a7=1000000007;
ll inf=100000000000000000;

ll min(ll a,ll b){
    if(a<b)return a;
    return b;
}
ll max(ll a,ll b){
    if(a>b)return a;
    return b;
}

int main(){
    ll n,m;cin>>n>>m;
    ll p[m],y[m];
    vector<vector<pair<ll,ll>>> aa(n+1); //県iの市の＜年、市の番号＞を入れる
    for(ll i=0;i<m;i++){
        cin>>p[i]>>y[i];
        aa[p[i]].push_back(make_pair(y[i],i));
    }
    string ans[m];
    for(ll i=1;i<=n;i++){
        sort(aa[i].begin(),aa[i].end());
        for(ll j=0;j<aa[i].size();j++){
            ans[aa[i][j].second]=to_string(j+1);
            while(ans[aa[i][j].second].size()<=6){
                ans[aa[i][j].second]='0'+ans[aa[i][j].second];
            }
            ans[aa[i][j].second]=to_string(i)+ans[aa[i][j].second];
            while(ans[aa[i][j].second].size()<=12){
                ans[aa[i][j].second]='0'+ans[aa[i][j].second];
            }
        }
    }
    for(ll i=0;i<m;i++){
        cout<<ans[i]<<endl;
    }
    
    return 0;
}