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 <bits/stdc++.h>
#define fort(i,n) for (int i = 1; i <= n; ++i)
#define ll long long
#define pi pair<int,int>
#define sz size()
#define er erase
#define fr first
#define sc second
using namespace std;
 
int n,m,r[100005],cnt[100005],a[100005],b[100005];
pair <pair<int,int>,int> p[100005];
 
int cif(int x)
{
    int t=0;
    while (x>0) x/=10, ++t;
    return t;
}
 
int main()
{
    cin >> n >> m;
    fort(i,m)
    {
        cin >> a[i] >> b[i];
        p[i].fr.sc=a[i], p[i].fr.fr=b[i], p[i].sc=i;
    }
    sort (p+1,p+m+1);
    fort(i,m)
        ++r[p[i].fr.sc], cnt[p[i].sc]=r[p[i].fr.sc];
    fort(i,m)
    {
        fort(j,6-cif(a[i]))
            cout << 0;
        cout << a[i];
        fort(j,6-cif(cnt[i]))
            cout << 0;
        cout << cnt[i] << '\n';
    }
}