#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
static mt19937 _g(time(nullptr));
inline long long int randint(long long int a, long long int b) {
  long long int w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
const int T = 666;
struct Seg {
  int l, r, idx, len;
};
bool cmp2(const Seg& lhs, const Seg& rhs) {
  if (lhs.len != rhs.len) return lhs.len < rhs.len;
  if (lhs.l != rhs.l) return lhs.l < rhs.l;
  return lhs.r < rhs.r;
}
void solve() {
  int n;
  cin >> n;
  vector<Seg> segs(n);
  for (auto& el : segs) cin >> el.l >> el.r;
  for (int i = 0; i < n; ++i)
    segs[i].idx = i, segs[i].len = segs[i].r - segs[i].l + 1;
  sort(segs.begin(), segs.end(), cmp2);
  vector<int> st;
  for (int i = 0; i < n;) {
    bool found = false;
    int x;
    for (int j = 0; j < T; ++j) {
      x = randint(segs[i].l, segs[i].r);
      bool used = false;
      for (auto y : st) used |= (x == y);
      if (!used) {
        found = true;
        break;
      }
    }
    if (!found) {
      st.pop_back();
      --i;
    } else {
      st.push_back(x);
      ++i;
    }
  }
  vector<int> ans(n);
  for (int i = 0; i < n; ++i) ans[segs[i].idx] = st[i];
  for (auto el : ans) cout << el << " ";
  cout << endl;
}
int main(int argc, char** argv) {
  fast_io();
  solve();
  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;
const long long int mod = 1e9 + 7;
static mt19937 _g(time(nullptr));
inline long long int randint(long long int a, long long int b) {
  long long int w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
const int T = 666;
struct Seg {
  int l, r, idx, len;
};
bool cmp2(const Seg& lhs, const Seg& rhs) {
  if (lhs.len != rhs.len) return lhs.len < rhs.len;
  if (lhs.l != rhs.l) return lhs.l < rhs.l;
  return lhs.r < rhs.r;
}
void solve() {
  int n;
  cin >> n;
  vector<Seg> segs(n);
  for (auto& el : segs) cin >> el.l >> el.r;
  for (int i = 0; i < n; ++i)
    segs[i].idx = i, segs[i].len = segs[i].r - segs[i].l + 1;
  sort(segs.begin(), segs.end(), cmp2);
  vector<int> st;
  for (int i = 0; i < n;) {
    bool found = false;
    int x;
    for (int j = 0; j < T; ++j) {
      x = randint(segs[i].l, segs[i].r);
      bool used = false;
      for (auto y : st) used |= (x == y);
      if (!used) {
        found = true;
        break;
      }
    }
    if (!found) {
      st.pop_back();
      --i;
    } else {
      st.push_back(x);
      ++i;
    }
  }
  vector<int> ans(n);
  for (int i = 0; i < n; ++i) ans[segs[i].idx] = st[i];
  for (auto el : ans) cout << el << " ";
  cout << endl;
}
int main(int argc, char** argv) {
  fast_io();
  solve();
  return 0;
}

```