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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int p[100010],y[100010];
int idx[100010];
int cmp(const void*a,const void*b){
	if(p[*(int*)a]<p[*(int*)b])return -1;
	if(p[*(int*)a]>p[*(int*)b])return  1;
	if(y[*(int*)a]<y[*(int*)b])return -1;
	return 1;
}

long long ans[100010];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",p+i,y+i);
		idx[i]=i;
	}
	qsort(idx,m,sizeof(int),cmp);
	int pre=-1,cnt=0;
	for(int i=0;i<m;i++){
		if(pre==p[idx[i]])cnt++;
		else{
			pre=p[idx[i]];
			cnt=1;
		}
		ans[idx[i]]=p[idx[i]]*1000000LL+cnt;
	}
	for(int i=0;i<m;i++)printf("%012lld\n",ans[i]);
}