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 ll = long long;
#define rep(i, begin, n) for (int i = begin; i < n; i++)
#define repe(i, begin, n) for (int i = begin; i <= n; i++)
#define repr(i, begin, n) for (int i = begin; i > begin - n; i--)
#define repre(i, begin, end) for (int i = begin; i >= end; 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;
}

const int inf = 1000000007;
const int MOD = 1000000007;
const long long INF = 1000000000000000007;

// -------------------------------------------------------

int N, M;
int P[101010], Y[101010];
string ids[101010];

string zero_padding(int n)
{
    string s = to_string(n);
    for (int i = 0; i < 10 - s.length(); i++)
    {
        s = '0' + s;
    }
    return s;
}

int main()
{
    cin >> N >> M;
    vector<vector<pair<int, int>>> PV(N + 1);
    repe(i, 1, M)
    {
        cin >> P[i] >> Y[i];
        PV[P[i]].push_back(make_pair(Y[i], i));
    }
    repe(i, 1, N)
    {
        sort(PV[i].begin(), PV[i].end());
        for (int j = 0; j < PV[i].size(); j++)
        {
            auto e = PV[i][j];
            string id = zero_padding(i) + zero_padding(j + 1);
            ids[e.second] = id;
        }
    }
    repe(i, 1, M)
    {
        cout << ids[i] << endl;
    }
}