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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lcm(long long x, long long y) {
  if (gcd(x, y) == 0ll) return 0;
  return x * y / gcd(x, y);
}
long long expmod(long long a, long long b) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y) % 1000000007;
    y = (y * y) % 1000000007;
    b >>= 1;
  }
  return (x) % 1000000007;
}
long long inv(long long n) { return expmod(n, 1000000007 - 2); }
map<int, int> m;
int main() {
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int> > v;
  int mn = 100000000;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(make_pair(a, b), i));
    mn = min(mn, a);
  }
  int visited[n], ans[n];
  memset(visited, 0, sizeof(visited));
  memset(ans, 0, sizeof(ans));
  sort(v.begin(), v.end());
  int flag = 0;
  while (flag == 0) {
    int fixed = 1000000000;
    for (int i = 0; i < n; i++) {
      int l = v[i].first.first;
      int r = v[i].first.second;
      if (visited[i] == 0 && mn >= l && mn <= r && r - l + 1 < fixed) {
        visited[i] = 1;
        ans[v[i].second] = mn;
        fixed = r - l + 1;
      }
    }
    mn++;
    for (int i = 0; i < n; i++) {
      int l = v[i].first.first;
      int r = v[i].first.second;
      if (visited[i] == 1) continue;
      if (mn >= l && mn <= r) {
        v[i].first.first = mn;
      }
    }
    sort(v.begin(), v.end());
    int flag2 = 0;
    for (int i = 0; i < n; i++) {
      int l = v[i].first.first;
      int r = v[i].first.second;
      if (visited[i] == 1) continue;
      if (mn >= l && mn <= r) {
        flag2 = 1;
        break;
      }
    }
    if (flag2 == 0)
      for (int i = 0; i < n; i++) {
        if (visited[i] == 0) {
          mn = max(v[i].first.first, mn);
          break;
        }
      }
    flag = 1;
    for (int i = 0; i < n; i++) {
      if (ans[i] == 0) {
        flag = 0;
        break;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}