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: Wrong Answer)

#include <iostream>
#include <vector>
#include <array>
#include <algorithm>
#include <string>
#include <cmath>
 
using namespace std;
 
typedef long long ll;
 
struct city {
	int ans1;
	int ans2;
};
 
int main() {
	ll N, M;
	cin >> N >> M;
	vector<ll> P(M), Y(M), ans(M), T(N, 1);
	vector<city> C(M);
 
	for (int i = 0; i < M; i++) {
		cin >> P[i] >> Y[i];
	}
 
	ll max = *(max_element(Y.begin(), Y.end()));
 
	//古いやつから消していく
	for (int i = 0; i < M; i++) {
		vector<ll>::iterator iter = min_element(Y.begin(), Y.end());
		int index = distance(Y.begin(), iter);
//		cout << "index " << index << " " << P[index] << " " << Y[index] << endl;
		C[index].ans1 = P[index];
//		cout << C[index].ans1 << endl;
		C[index].ans2 = T[P[index]];
		T[(P[index])]++;
		Y[index] = max + 1;
	}
 
	for (int i = 0; i < M; i++) {
		printf("%06d%06d\n", C[i].ans1, C[i].ans2);
	}
}