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;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const ll INF = 1LL<<60;
const ll MOD = 1e9 + 7;
// Driver function to sort the vector elements 
// by second element of pairs 
bool sortbysec(const pair<pair<ll,ll>,ll> &a, 
              const pair<pair<ll,ll>,ll> &b) 
{ 
    return (a.second < b.second); 
} 

void ans(ll i, ll j) {
    cout << setfill('0') << right << setw(6) << i; 
    cout << setfill('0') << right << setw(6) << j << endl;

}

int main() {
    ll N,M;
    cin >> N >> M;
    vector<pair<pair<ll,ll>,ll>> vec;
    for (ll i = 0;i<M;i++) {
        ll p,y;
        cin >> p >> y;
        pair<pair<ll,ll>,ll> pp;
        pp = make_pair(make_pair(p,y),i);
        vec.push_back(pp);
    }
    sort(vec.begin(),vec.end());
    ll city = vec.at(0).first.first; //初めの市
    ll cnt = 1;
    for (ll i = 0;i<M;i++) {
        if (vec.at(i).first.first == city) {
            vec.at(i).first.second = cnt;
            cnt++;
        }
        else {
            cnt = 1;
            city = vec.at(i).first.first;
            vec.at(i).first.second = 1;
        }
    }
    sort(vec.begin(),vec.end(),sortbysec);

    for (ll i = 0;i<M;i++) {
        ans(vec.at(i).first.first,vec.at(i).first.second);
    }
}