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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define rep(i,n) for (int i = (0); i < (n); ++i)
#define rng(x) (x).begin(),(x).end()
#define rrng(x) (x).rbegin(),(x).rend()
#define debug(x) cout << #x << " = " << (x) << endl;
#define SUM(x) accumulate(rng(x),0)
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define uni(x) x.erase(unique(rng(x)),x.end())
#define SP << " " <<
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using pii = pair<int, int>;
using tiii = tuple<int, int, int>;
const int mod = 1000000007;
const double EPS = 1e-9;
const int INF = 1<<30;
const ll INFLL = 1LL<<60;
//cout << fixed << setprecision(10);

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main () {
  int n, m;
  cin >> n >> m;
  vector<tiii> data(m);
  rep(i,m) {
    int p, y;
    cin >> p >> y;
    data[i] = tiii(p,y,i);
  }
  sort(data.begin(), data.end());
  int x;
  int prev_p = get<0>(data[0]);
  vector<tiii> id(m);
  rep(i,m) {
    int p, y, num;
    tie(p, y, num) = data[i];
    if (prev_p != p) x = 1;
    else x++;
    id[i] = tiii(num, p, x);
  }
  sort(id.begin(), id.end());
  rep(i,m) printf("%06d%06d\n", get<1>(id[i]), get<2>(id[i]));
  return 0;
}