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<bits/stdc++.h>
using namespace std;

int i1,i2;char ch;
inline int read()
{
	i1=i2=ch=0;
	while(!isdigit(ch))	i1|=ch=='-',ch=getchar();
	while(isdigit(ch))	i2=(i2<<1)+(i2<<3)+(ch^48),ch=getchar();
	return i1? -i2:i2;
}

int n,m;
#define mp make_pair
#define pii pair<int,int>
struct node
{
	int rankk,id,bel,y;
}e[100010];

bool cmp1(node a,node b) {return a.bel<b.bel||(a.bel==b.bel&&a.y<b.y);}
bool cmp2(node a,node b) {return a.id<b.id;}

int dig(int x)
{
	int tot=0;
	while(x)
	tot++,x/=10;
	return tot;
}

int main()
{
	n=read(),m=read();
	for(int i=1;i<=m;i++)
	{
		e[i].bel=read(),e[i].y=read();
		e[i].id=i;
	}
	
	sort(e+1,e+1+m,cmp1);
	int tot=0;
	for(int i=1;i<=m;i++)
	{
		if(e[i].bel^e[i-1].bel) tot=0;
		e[i].rankk=++tot;
	}
	
	sort(e+1,e+1+m,cmp2);
	
	int tmp1,tmp2;
	for(int i=1;i<=m;i++)
	{
		tmp1=6-dig(e[i].bel),tmp2=6-dig(e[i].rankk);
		for(int j=1;j<=tmp1;j++) putchar('0');
		printf("%d",e[i].bel);
		for(int j=1;j<=tmp2;j++) putchar('0');
		printf("%d\n",e[i].rankk);
	}
	
	return 0;
}