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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <map>
#include <vector>
#include <utility>
using namespace std;

void opt(int p, int y) {
    if (10000 <= p && p < 100000) cout << "0" << p;
    else if  (1000 <= p && p < 10000) cout << "00" << p;
    else if  (100 <= p && p < 1000) cout << "000" << p;
    else if  (10 <= p && p < 100) cout << "0000" << p;
    else if  (1 <= p && p < 10) cout << "00000" << p;
    else cout << p;

    if (10000 <= y && y < 100000) cout << "0" << y;
    else if  (1000 <= y && y < 10000) cout << "00" << y;
    else if  (100 <= y && y < 1000) cout << "000" << y;
    else if  (10 <= y && y < 100) cout << "0000" << y;
    else if  (1 <= y && y < 10) cout << "00000" << y;
    else cout << y;

    cout << endl;
}
int main(void) {
    int N,M;cin>>N>>M;
    // memo(出力のときに使う)
    vector<pair<int, int> > memo;

    map<int, map<int, int> > mp;
    for (int i = 0; i < M; i++) {
        int P,Y;cin>>P>>Y;
        pair<int, int> p;
        p.first = P;
        p.second = Y;
        memo.push_back(p);
        if (mp.count(P) > 0) {
            map<int, int> m = mp[P];
            m[Y] = 1; // temporary
            mp[P] = m;
        } else {
            map<int, int> m;
            m[Y] = 1;
            mp[P] = m;
        }
    }
    // sort phase
    map<int, map<int, int> > newMap;
    for (map<int, map<int, int> >::iterator it = mp.begin(); it != mp.end(); ++it) {
        map<int, int> m = it->second;
        int juni = 1;
        map<int, int> newMap1;
        for (map<int, int>::iterator itt = m.begin(); itt != m.end(); ++itt) {
            newMap1[itt->first] = juni;
            juni++;
        }
        newMap[it->first] = newMap1;
    }

    // output phase
    for (int i = 0; i < M; i++) {
        pair<int, int> pr = memo.at(i);
        map<int, int> m = newMap[pr.first];
        opt(pr.first, m[pr.second]);
    }
    return 0;
}