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: Wrong Answer)

#include<stdio.h>

int main(){
	int N , M;
	scanf("%d%d",&N,&M);
	int data[M][4];
	for(int i=0;i<M;i++){
		data[i][0] = i;
		scanf("%d%d",&data[i][1],&data[i][2]);
	}
	int med3(int x, int y, int z) {
		if (x < y) {
			if (y < z) return y; else if (z < x) return x; else return z;
		} else {
			if (z < y) return y; else if (x < z) return x; else return z;
		}
	}

	void quicksort(int a[][4], int left, int right) {
		if (left < right) {
			int i = left, j = right;
			int tmp, pivot = med3(a[i][2], a[i + (j - i) / 2][2], a[j][2]);
			while (1) {
				while (a[i][2] < pivot) i++;
				while (pivot < a[j][2]) j--;
				if (i >= j) break;
				tmp = a[i][0]; a[i][0] = a[j][0]; a[j][0] = tmp;
				tmp = a[i][1]; a[i][1] = a[j][1]; a[j][1] = tmp;
				tmp = a[i][2]; a[i][2] = a[j][2]; a[j][2] = tmp;
				i++; j--;
			}
			quicksort(a, left, i - 1);
			quicksort(a, j + 1, right);
		}
	}
	quicksort(data,0,M-1);
	int ansp[M];
	int ansy[M];
	int index[N];
	for(int k=0;k<M;k++){
		int num = data[k][0];
		int pre = data[k][1];
		ansp[num] = pre;
		index[pre]++;
		ansy[num] = index[pre];
	}
	for(int k=0;k<M;k++){
		printf("%06d%06d\n",ansp[k],ansy[k]);
	}
}