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 <bits/stdc++.h>
 
using namespace std;
#define all(x) (x).begin(),(x).end()
 
 
void solve() {
    int N, M;
    cin >> N >> M;
    vector<int> P(M), Y(M);
    vector<vector<int>> data(N + 1);
    for (int i = 0; i < M; i++) {
        cin >> P[i] >> Y[i];
        data[P[i]].push_back(Y[i]);
    }
 
    for (int i = 0; i < N + 1; i++) {
        sort(all(data[i]));
    }
 
    for (int i = 0; i < M; i++) {
        vector<int> &d = data[P[i]];
        int k = lower_bound(all(d), Y[i]) - d.begin();
        
        printf("%06d%06d\n", P[i], k + 1);
    }
 
}
 
int main() {
    solve();
    return 0;
}