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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, a, b) for(int i = a; i < b; i++)
#define  all(x) (x).begin(),(x).end()     // 昇順ソート
#define  rall(v) (v).rbegin(), (v).rend() // 降順ソート
#define  FastIO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef long long int lli;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return true; } return false; }

int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};


int main(){
    int n, m;
    cin >> n >> m;
    vector<vector<pair<int, int>>> c(n);
    vector<pair<int, int>> ans(m);
    rep(i, m){
        int pi, year;
        cin >> pi >> year;
        pi--;
        c[pi].push_back(make_pair(year, i));
    }
    rep(i,n){
        sort(all(c[i]));
        rep(j, c[i].size()){
            int id = c[i][j].second;
            ans[id] = make_pair(i+1, j+1);
        }
    }
    rep(i,m){
        printf("%06d%06d\n", ans[i].first, ans[i].second);
    }
    return 0;
}