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 <vector>
#include <algorithm>
#include <map>
#include <cstring>
#include <math.h>
#include <bitset>
#include <queue>
#include <set>
#include <iomanip>
#include <math.h>
#include <assert.h>
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr long long int INFLL = 1001001001001001LL;
constexpr long long int infll = 1001001001001001LL;
constexpr int INF = 1000000007;
constexpr int inf = 1000000007;
const int mod = 1000000007;

template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

ll gcd(ll a, ll b)
{
    if (a % b == 0)
    {
        return (b);
    }
    else
    {
        return (gcd(b, a % b));
    }
}

ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}

inline double nCr(const int n, int r)
{
    if (n == 0)
    {
        return 0;
    }
    if (r == 0)
    {
        return 1;
    }
    if (r == 1)
    {
        return n;
    }
    if (n == r)
    {
        return 1;
    }

    if (r > n / 2)
    {
        r = n / 2;
    }

    double result = 1;
    for (double i = 1; i <= r; i++)
    {
        result *= (n - i + 1) / i;
    }

    return (result);
}

template <typename T>
T seinomi(T a)
{
    if (a > 0)
    {
        return a;
    }
    else
    {
        return 0;
    }
}

template <typename T>
map<T, T> soinsuubunkai(T n) //連想配列[素因数f.first][個数f.second]
{
    map<T, T> ret;
    for (T i = 2; i * i <= n; i++)
    {
        while (n % i == 0)
        {
            ret[i]++;
            n /= i;
        }
    }
    if (n != 1)
        ret[n] = 1;
    return ret;
}

template <typename T>
T ketasuu(T num)
{
    return std::to_string(num).length();
}

int kaizyou(int k)
{
    int sum = 1;
    for (int i = 1; i <= k; ++i)
    {
        sum *= i;
    }
    return sum;
}

template <class ForwardIt, class T>
void iota(ForwardIt first, ForwardIt last, T value)
{
    while (first != last)
    {
        *first++ = value;
        ++value;
    }
}

//nの階乗を計算
int factorial(int n)
{
    int answer = 1;
    while (n > 1)
    {
        answer *= n;
        n--;
    }
    return answer;
}
int comb(int n, int r)
{
    return factorial(n) / (factorial(r) * factorial(n - r));
}

long long modpow(long long a, long long n, long long mod)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

//小数点以下10桁テンプレート（main関数内の最初に貼付け）
//std::cout << std::fixed << std::setprecision(10);

//----------------------------------------------------------------
int main()
{
    ll n, m;
    cin >> n >> m;
    ll y, p;

    vector<vector<ll>> a(100010, vector<ll>());
    map<ll, ll> b, c, d;

    for (int i = 0; i < m; i++)
    {
        cin >> p >> y;
        a[p].push_back(y);
        b[y] = p;
        c[i] = y;
    }
    for (int i = 1; i <= n; i++)
    {
        sort(a[i].begin(), a[i].end());
    }
    for (int i = 1; i <= n; i++)
    {
        ll cnt = 1;
        for (auto f : a[i])
        {
            d[f] = cnt;
            cnt++;
        }
    }
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < 6 - ketasuu(b[c[i]]); j++)
        {
            cout << "0";
        }
        cout << b[c[i]];
        for (int j = 0; j < 6 - ketasuu(d[c[i]]); j++)
        {
            cout << "0";
        }
        cout << d[c[i]] << endl;
    }
}