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: Accepted)

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

# define REP(i,n) for (int i=0;i<(n);++i)
# define rep(i,a,b) for(int i=a;i<(b);++i)
# define p(s) std::cout << s ;
# define pl(s)  std::cout << s << endl;
# define printIf(j,s1,s2) cout << (j ? s1 : s2) << endl;
# define YES(j) cout << (j ? "YES" : "NO") << endl;
# define Yes(j) std::cout << (j ? "Yes" : "No") << endl;
# define yes(j) std::cout << (j ? "yes" : "no") << endl;
# define all(v) v.begin(),v.end()
template<class T> inline bool chmin(T &a, T b){ if(a > b) { a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b){ if(a < b) { a = b; return true;} return false;}
typedef long long int ll;

const ll longinf=1LL<<60 ;

vector<int> g[100005];

int main() {
    int N, M;
    cin >> N >> M;

    vector<int> P(M), Y(M);
    REP(i, M) {
        cin >> P[i] >> Y[i];
        g[P[i]].push_back(Y[i]);
    }
    REP(i, N) if(!g[i + 1].empty()) sort(all(g[i + 1]));

    REP(i, M) {
        int d = (upper_bound(all(g[P[i]]), Y[i]) - g[P[i]].begin());
        printf("%06d%06d\n", P[i], d);
    }

    return 0;
}