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<bits/stdc++.h>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62)
#define PI 3.14159265358979

typedef long long i64;
typedef pair<i64,i64> P;

inline i64 mod(i64 a, i64 m) { return (a % m + m) % m; }

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }



int n, m, a[101010], b[101010];
vector<P> v[101010];
string ans[101010];
void solve(){
	cin >> n >> m;
	for(int i = 0; i < m; i++){
		cin >> a[i] >> b[i];
		v[a[i]].push_back({b[i],i});
	}
	for(int i = 0; i <= n; i++){
		sort(v[i].begin(),v[i].end());
		int cnt = 1;
		for(auto j : v[i]){
			int x = j.second;
			for(int k = 100000; k > 0; k/=10){
				ans[x] += (char)('0'+a[x]/k);
			}
			for(int k = 100000; k > 0; k/=10){
				ans[x] += (char)('0'+cnt/k);
			}
			cnt++;
		}
	}
	
	for(int i = 0; i < m; i++){
		cout << ans[i] << endl;
	}
}

int main(){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	int t = 1;
	//cin >> t;
	while(t--){
		solve();
	}
	
	return 0;
}