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;

using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)

int main()
{
    lint n,m;cin >> n >>m;
    vector<vector<int>> v(m,vector<int>(3));
    REP(i, m) {
        int p, y;
        cin >> v[i][0] >> v[i][1];
        v[i][2] = i;
        }
    lint ans=0;
    vector<vector<int>> number(m, vector<int>(2));
    sort(ALL(v));
    int j = 1,c=1;
    REP(i,m){
        if (v[i][0] != j){

            j++;
            c= 1;
        }
        number[v[i][2]][0] = v[i][0];
        number[v[i][2]][1] = c;
        c++;
    }

    REP(i, m){
       // cout << number[i][0] <<" "<< number[i][1]<<"\n";
       cout << std::setfill('0') << std::right << std::setw(6) << number[i][0];
       cout << std::setfill('0') << std::right << std::setw(6) << number[i][1] << "\n";
    }

    REP(i, m)
    {
        //cout << v[i][0] <<" "<< v[i][1] <<" "<< v[i][2] << "\n";
    }

    //cout << ans <<"\n";

    return 0;
}