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>
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define rep(i,n) for(ll i=0, i##_len=(n); i<i##_len; ++i)
using namespace std;
#define sz(x) ((int)(x).size())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MEMSET(v, h) memset((v), h, sizeof(v))
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; }
#define pb push_back
#define mp make_pair
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
const ll INF = 1LL<<60;
 
int main() {
	int N, M;
	cin >> N >> M;
 
	vector<tuple<ll, ll, ll>> data(M);
 
	for (ll i = 0; i < M; ++i) {
		ll p, y;
		cin >> p >> y;
		data[i] = make_tuple(y, p, i);
	}
 
 
	sort(data.begin(), data.end());
 
	vector<ll> cnt(N + 1);
 
 
	vector<pair<ll, ll>> id(M);
 
	for (ll i = 0; i < M; ++i) {
		ll y, p, idx;
		tie(y, p, idx) = data[i];
		++cnt[p];
		id[idx] = make_pair(p, cnt[p]);
	}
 
	for (ll i = 0; i < M; ++i) {
		cout << setw(6) << setfill('0') << id[i].first;
		cout << setw(6) << setfill('0') << id[i].second << endl;
	}
	return 0;
}