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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cmath>
#include <iomanip>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <bitset>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll INF = 1e18;
constexpr ll MOD = 1e9 + 7;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m, p[112345], y[112345];
    vector<P> q[112345];
    cin >> n >> m;
    vector<P> res(m);
    REP(i, m)
    {
        cin >> p[i] >> y[i];
        q[p[i]].push_back(make_pair(y[i], i));
    }

    REP(i, n + 1)
    {
        if (q[i].size())
        {
            SORT(q[i]);
            int cnt = 1;
            REP(j, q[i].size())
            {
                res[q[i][j].second].first = p[q[i][j].second];
                res[q[i][j].second].second = cnt;
                cnt++;
            }
        }
    }

    REP(i, m)
    {
        printf("%06d%06d\n", res[i].first, res[i].second);
    }

    return 0;
}