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: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>
#include <time.h>

using namespace std;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, M;
	cin >> N >> M;

	vector<vector<int>> p;
	vector<int> g;
	g.resize(M + 1);
	p.resize(3);
	p.resize(M);
	p.resize(M);
	p.resize(M);
	;
	for (int m = 0; m < M; ++m) {
		int k1, k2;
		cin >> k1 >> k2;
		p[m].resize(3);
		p[m][0] = k1;
		p[m][1] = k2;
		p[m][2] = m;
		g[m] = 1;
	}

	clock_t start = clock();
	vector<string> s;
	s.resize(M + 10);

	sort(p.begin(), p.end());


	for (int m = 0; m < M; ++m) {
		char c[50];
		sprintf(c, "%06d%06d", p[m][0], g[p[m][0]]);
		++g[p[m][0]];
		s[p[m][2]] = c;
	}
	for (int m = 0; m < M; ++m) {
		cout << s[m].c_str()　<<endl;
	}

//	clock_t end = clock();
//	const double time = static_cast<double>(end - start) / CLOCKS_PER_SEC * 1000.0;
//	printf("time %lf[ms]\n", time);
//	system("pause");
	return 0;
}