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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;

template<class T> inline void chmin(T& a, T b) {
    if (a > b) {
        a = b;
    }
}

template<class T> inline void chmax(T& a, T b) {
    if (a < b) {
        a = b;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);std::cout<<std::fixed<<std::setprecision(10);
    int N,M;
    cin >> N >> M;
    vector<pair<int,int>> city(M);
    vector<vector<int>> Pre(N,vector<int>(0));
    REP(i,M){
        cin >> city[i].first;
        cin >> city[i].second;
        city[i].first--;
        Pre[city[i].first].push_back(city[i].second);
    }
    REP(i,N)sort(Pre[i].begin(),Pre[i].end());
    REP(i,M){
        for(int l = 0;;l++){
            if(city[i].second == Pre[city[i].first][l]){
                city[i].second = l+1;
                break;
            }
        }
    }
    REP(i,M){
        REP(l,6 - to_string(city[i].first+1).size())cout << 0;
        cout << city[i].first + 1;
        REP(l,6 - to_string(city[i].second).size())cout << 0;
        cout << city[i].second<< "\n";
    }
}