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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
typedef long long int ll;
#define mod 1000000007
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define m0(x) memset(x,0,sizeof(x))

int main(void){
	int n,m;
	scanf("%d %d",&n,&m);
	ll p[m],y[m],rank;
	rep(i,m){
		scanf("%lld %lld",&p[i],&y[i]);
	}
	rep(i,m){
		rank=1;
		printf("%06lld",p[i]);
		rep(j,m){
			if(p[j]==p[i]){
				if(y[j]<y[i]){
					rank++;
				}
			}
		}
		printf("%06lld\n",rank);
	}
	return 0;
}