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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <cstdlib>

using namespace std;
typedef long long int ll;
typedef pair<ll, ll> pi;

const ll MOD = 1e9+7;

static long gcd(long a, long b) {
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}
/*
long long int facctorialMethod(int k){//階乗
    int sum = 1;
    for (int i = 1; i <= k; ++i)
    {
        sum *= i;
    }
    return sum;
}*/

long long int tousasum(long long int n){//等差数列
    return n*(n+1)/2;
}

unsigned GetDigit(unsigned num){//桁数
    unsigned digit=0;
    while(num!=0){
        num /= 10;
        digit++;
    }
    return digit;
}

bool IsPrime(int num)//素数判定
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}

map< ll, ll > prime_factor(ll n) {//素因数分解
    map< ll, ll > ret;
    for(ll i = 2; i * i <= n; i++) {
        while(n % i == 0) {
            ret[i]++;
            n /= i;
        }
    }
    if(n != 1) ret[n] = 1;
    return ret;
}

ll calcNumOfCombination(ll n, ll r){
    ll num = 1;
    for(ll i = 1; i <= r; i++){
        num = num * (n - i + 1) / i;
    }
    return num;
}

int bubblesort(vector<int> array,int size){//バブルソート　転倒数
    int kaisu = 0;
    for(int i = 0; i < size; i++){
        for(int j = i + 1; j < size; j++){
            if(array[i] > array[j]){
                int number = array[i];
                array[i] = array[j];
                array[j] = number;
                kaisu++;
            }
        }
    }
    return kaisu;
}

ll facctorialMethod(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum = (sum * i)%MOD;
    }
    return sum;
}

int main() {
    ll N,M;
    cin >> N >> M;
    ll p[M],y[M];
    vector<ll> list[100001];

    for(int i=0;i<M;i++){
        cin >> p[i] >> y[i];
        list[p[i]].push_back(y[i]);
    }

    for(int i=0;i<=N;i++){

            sort((list[i]).begin(),(list[i]).end());



    }


    for(int i=0;i<M;i++){
        auto it = lower_bound(list[p[i]].begin(), list[p[i]].end(), y[i]);

        int index = it - list[p[i]].begin();

        printf("%06lld",p[i]);
        printf("%06lld\n",index+1);
    }

}