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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> P(m), Y(m);
    for(int i=0; i<m; ++i) cin >> P[i] >> Y[i], --P[i];

    vector<vector<int>> vals(n);
    for(int i=0; i<m; ++i) vals[P[i]].push_back(Y[i]);

    for(int v=0; v<n; ++v) {
        sort(vals[v].begin(), vals[v].end());

        // 今回は不要だが、通常は数のダブリをなくす
        vals[v].erase(unique(vals[v].begin(), vals[v].end()), vals[v].end());
    }

    for(int i=0; i<m; ++i) {
        int v = P[i];
        printf("%06d", v+1);

        // 座標圧縮して導いた答え
        int id = lower_bound(vals[v].begin(), vals[v].end(), Y[i]) - vals[v].begin();
        printf("%06d\n", id+1);
    }

}