#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  set<pair<pair<int, int>, int>> sp;
  int res[n];
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    sp.insert({{r - l + 1, l}, i});
  }
  set<int> si;
  for (auto it : sp) {
    int d;
    if (it.second % 2 == 0) {
      d = 1;
      res[it.second] = it.first.second;
    } else {
      d = -1;
      res[it.second] = it.first.second + it.first.first - 1;
    }
    while (si.count(res[it.second])) {
      res[it.second] += d;
    }
    si.insert(res[it.second]);
  }
  for (int i = 0; i < n; i++) cout << res[i] << " ";
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.

Output

Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.

Examples

Input

3
1 2
2 3
3 4


Output

1 2 3 


Input

2
1 3
1 3


Output

1 2 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  set<pair<pair<int, int>, int>> sp;
  int res[n];
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    sp.insert({{r - l + 1, l}, i});
  }
  set<int> si;
  for (auto it : sp) {
    int d;
    if (it.second % 2 == 0) {
      d = 1;
      res[it.second] = it.first.second;
    } else {
      d = -1;
      res[it.second] = it.first.second + it.first.first - 1;
    }
    while (si.count(res[it.second])) {
      res[it.second] += d;
    }
    si.insert(res[it.second]);
  }
  for (int i = 0; i < n; i++) cout << res[i] << " ";
  return 0;
}

```