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: Runtime Error)

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

int compare_int(const void *a, const void *b)
{
    return *(long*)a - *(long*)b;
}


int main(){
	int n,m,p[100000],i,k,j,ban[100000]={0};
	int ans[100000][2];
	scanf("%d%d",&n,&m);
	long y[n][100000],ycpy[100000];	
	for(i=0;i<m;i++){
		scanf("%d",&p[i]);
		scanf("%ld",&y[p[i]-1][ban[p[i]-1]]);
		ans[i][0]=p[i];
		ycpy[i]=y[p[i]-1][ban[p[i]-1]];
		ban[p[i]-1]++;
/*		if(now>y[i])now=y[i];
*/	}
	for(i=0;i<n;i++)qsort(y[i],ban[i],sizeof(long),compare_int);	
	puts("OK");
	for(j=0;j<n;j++){
//		printf("%ld\n",y[i]);
		for(k=0;k<ban[j];k++){
		printf("%ld\n",y[j][k]);
		for(i=0;i<m;i++){
			if(ycpy[i]==y[j][k]){
				ans[i][1]=k+1;
			}
		}
		}
	}
	for(i=0;i<m;i++)printf("%06d%06d\n",ans[i][0],ans[i][1]);
	return 0;
}