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 LL long long

vector<pair<int,int> > G[100009];
string ans[100009];

int get_len(int x){
	int val=0;
	while(x>0){
		val++;
		x/=10;
	}
	return val;
}

int main(){
	

	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n,m;
	cin>>n>>m;

	for(int i=1;i<=m;i++){
		int x,y;
		cin>>x>>y;

		G[x].push_back(make_pair(y,i));
	}
	

	for(int i=1;i<=n;i++){
		sort(G[i].begin(),G[i].end());
	}
	
	for(int i=1;i<=n;i++){
		for(int j=0;j<(int)G[i].size();j++){
			
			int idx = G[i][j].second;
			string res;
			int len = get_len(i);
			
			while(len<6){
				res.push_back('0');
				len++;
			}
			stack<int> st;
			int x=i;
			while(x>0){
				st.push(x%10);
				x/=10;
			}
			while(!st.empty()){
				res.push_back(st.top()+'0');st.pop();
			}

			len = get_len(j+1);
			while(len<6){
				res.push_back('0');
				len++;
			}
			x=j+1;
			while(x>0){
				st.push(x%10);
				x/=10;
			}
			while(!st.empty()){
				res.push_back(st.top()+'0');st.pop();
			}
			ans[idx] = res;
		}
	}
	for(int i=1;i<=m;i++){cout<<ans[i]<<endl;}
}