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: Time Limit Exceeded)

#include <stdio.h>
#include<stdlib.h>
// void MergeSort(int x[ ], int left, int right);
int main(){
  int n,m,i,j,k;
  scanf("%d %d",&n,&m);
  int p[m],y[m],x[m],yet[m];
  for(i=0;i<m;i++){
    scanf("%d %d",&p[i],&y[i]);
    x[i]=1;
    yet[i]=0;
  }

  int count,same[n],temp;
  for(j=0;j<m;j++){
    if(yet[j]==0){
      yet[j]=1;
      count = 0;
      same[0]=j;
      for(i=j+1;i<m;i++){
        if(p[i]==p[j]){
          yet[i]=1;
          count+=1;
          same[count]=i;
        }
      }

      if(count>0){
        for(i=0;i<count+1;i++){
          for(k=count;k>i;k--){
            if(y[same[k]]<y[same[k-1]]){
              temp=same[k];
              same[k]=same[k-1];
              same[k-1]=temp;
            }
          }
        }
        for(i=0;i<count+1;i++){
          x[same[i]]=i+1;
        }
      }


    }
  }

  for(i=0;i<m;i++){
    printf("%06d%06d\n",p[i],x[i]);
  }
  return 0;
}
//
// void MergeSort(int x[ ], int left, int right)
// {
//     int mid, i, j, k;
//
//     if (left >= right)              /* 配列の要素がひとつなら */
//         return;                     /* 何もしないで戻る */
//
//                                     /* ここでは分割しているだけ */
//     mid = (left + right) / 2;       /* 中央の値より */
//     MergeSort(x, left, mid);        /* 左を再帰呼び出し */
//     MergeSort(x, mid + 1, right);   /* 右を再帰呼び出し */
//
//       /* x[left] から x[mid] を作業領域にコピー */
//     for (i = left; i <= mid; i++)
//         temp[i] = x[i];
//
//       /* x[mid + 1] から x[right] は逆順にコピー */
//     for (i = mid + 1, j = right; i <= right; i++, j--)
//         temp[i] = x[j];
//
//     i = left;         /* i とj は作業領域のデーターを */
//     j = right;        /* k は配列の要素を指している */
//
//     for (k = left; k <= right; k++)    /* 小さい方から配列に戻す */
//         if (temp[i] <= temp[j])        /* ここでソートされる */
//             x[k] = temp[i++];
//         else
//             x[k] = temp[j--];
// }