Queue CodeForces - 141C

In the Main Berland Bank n people stand in a queue at the cashier, everyone knows his/her height h_i, and the heights of the other people in the queue. Each of them keeps in mind number a_i — how many people who are taller than him/her and stand in queue in front of him.

After a while the cashier has a lunch break and the people in the queue seat on the chairs in the waiting room in a random order.

When the lunch break was over, it turned out that nobody can remember the exact order of the people in the queue, but everyone remembers his number a_i.

Your task is to restore the order in which the people stood in the queue if it is possible. There may be several acceptable orders, but you need to find any of them. Also, you need to print a possible set of numbers h_i — the heights of people in the queue, so that the numbers a_i are correct.

Input

The first input line contains integer n — the number of people in the queue (1 ≤ n ≤ 3000). Then n lines contain descriptions of the people as "name_i a_i" (one description on one line), where name_i is a non-empty string consisting of lowercase Latin letters whose length does not exceed 10 characters (the i-th person's name),a_i is an integer (0 ≤ a_i ≤ n - 1), that represents the number of people who are higher and stand in the queue in front of person i. It is guaranteed that all names are different.

Output

If there's no acceptable order of the people in the queue, print the single line containing "-1" without the quotes. Otherwise, print in n lines the people as "name_ih_i", where h_i is the integer from 1 to 10⁹ (inclusive), the possible height of a man whose name is name_i. Print the people in the order in which they stand in the queue, starting from the head of the queue and moving to its tail. Numbers h_i are not necessarily unique.

Example

Input

4
a 0
b 2
c 0
d 0

Output

a 150
c 170
d 180
b 160

Input

4
vasya 0
petya 1
manya 3
dunay 3

Output

-1

思路： 設最低的高度爲1，先把們每個人按照a從小到大排序，那麼第i個人前面有a個人比他高，那麼就使這個人的高度爲i-a，那麼爲了滿足a，使這個人前面高度>=i-a的人的高度++，然後就可以了.

至於什麼時候無法滿足，當i-1 < a的時候輸出-1.

至於爲什麼這麼做呢，手動模擬一下你就知道其中的奧祕，多說無益，自己在草稿紙上寫一下.

上代碼:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
  string name;
  int a;
  int h;
};
node op[3010];
bool cmp(node x,node y)
{
    return x.a < y.a;
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        for(int i=1; i<=n; i++)
        {
          cin >> op[i].name >> op[i].a;
          op[i].h = 0;
        }
        sort(op+1,  op+1+n, cmp);
        int flag = 1;
        if(op[1].a!=0)flag = 0;
        if(flag)
        for(int i=1; i<=n; i++)
        {
           if(i-1 < op[i].a){flag=0; break;}
           op[i].h = i-op[i].a;
           for(int k=1; k<i; k++)
           {
               if(op[k].h >= op[i].h)++op[k].h;
           }
        }
        if(flag==0)
            printf("-1\n");
        else
          for(int i=1; i<=n; i++)
          {
              cout << op[i].name << " " << op[i].h << endl;
          }
    }
    return 0;
}

水波.