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<algorithm>
using namespace std;
void pz(int n){
	for(int i=0;i<n;i++){
		cout<<'0';
	}
	return;
}
void pl(int n){
	int cnt=0;
	int m=n;
	while (n>0){
		cnt++;
		n=n/10;
	}
	pz(6-cnt);
	if(m!=0) cout<<m;
	return;
}
int search(vector<long long int> a,long long int n){//a[ans]<=n and ans is the biggest
	if (a[0]>n) return -1;
	int ok=0,ng=a.size(),mid;
	while(ng-ok>1){
		mid=(ok+ng)/2;
		if(a[mid]<=n){
			ok=mid;
		}else{
			ng=mid;
		}
	}
	return ok;
}


int main(){
	int n,m;
	cin>>n>>m;
	vector<int> p(m),y(m);
	long long int a;
	vector<long long int> ho(m);
	for(int i=0;i<m;i++){
		cin>>p[i];
		cin>>y[i];
		ho[i]=p[i]*1e6+y[i];
	}
	sort(ho.begin(),ho.end());
	/*for(int i=0;i<ho.size();i++){
		cout<<ho[i]<<" ";
	}
	cout<<endl;*/
	for(int i=0;i<m;i++){
		a=p[i]*1e6+y[i];
		//cout<<"#"<<a<<" "<<a-y[i]<<endl;
		//cout<<"!"<<search(ho,a)<<" "<<search(ho,a-y[i])<<endl;
		pl(p[i]);
		pl(search(ho,a)-search(ho,a-y[i]));
		cout<<endl;
	}
	return 0;
}