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>

typedef struct {
	int No;//通しナンバー
	int P;
	int Y;
	int ans;//得たい値である県ごとの順番
}CITY;

int Comp1(const void* pElem1, const void* pElem2)
{
	// データが扱いやすいので一旦変数に入れておく
	const CITY* pmem1 = (const CITY*)pElem1;
	const CITY* pmem2 = (const CITY*)pElem2;
	int nDiff;  // 両要素の差
	nDiff = pmem1->P - pmem2->P;
	if (nDiff != 0)
		return nDiff;
	nDiff = pmem1->Y - pmem2->Y;
	if (nDiff != 0)
		return nDiff;
}

int Comp2(const void* pElem1, const void* pElem2)
{
	// データが扱いやすいので一旦変数に入れておく
	const CITY* pmem1 = (const CITY*)pElem1;
	const CITY* pmem2 = (const CITY*)pElem2;
	int nDiff;  // 両要素の差
	nDiff = pmem1->No - pmem2->No;
	if (nDiff != 0)
		return nDiff;
}

CITY city[10002];//構造体配列を宣言

int main() {
	int N, M;
	scanf("%d%d", &N, &M);
	//CITY city[10002];//構造体配列を宣言
	for (int i = 0; i < M; i++)
	{
		city[i].No = i;
		scanf("%d%d", &city[i].P, &city[i].Y);
		//得たい値である県ごとの市の順番
			//-1の場合		:まだ未確定(初期化)
			//0以上の場合	:確定
		city[i].ans = -1;
	}

	//県順誕生年順にソートする
	qsort(city, M, sizeof(CITY), Comp1);

	for (int i = 0; i < M; i++) {
		if (i == 0) {
			city[i].ans = 1;
		}
		else if (i > 0 && city[i].P != city[i - 1].P) {
			city[i].ans = 1;
		}
		else {
			city[i].ans = city[i - 1].ans + 1;
		}
	}

	//通しナンバー順にソートする
	qsort(city, M, sizeof(CITY), Comp2);

	//printf("ソート結果\n");
	//for (int i = 0; i < M; i++) {
	//	printf("%d %d %d %d\n", city[i].No, city[i].P, city[i].Y, city[i].ans);
	//}
	for (int i = 0; i < M; i++) {
		printf("%06d%06d\n", city[i].P, city[i].ans);
	}
	return 0;
}