#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
pair<pair<long long, long long>, long long> a[100005];
pair<pair<long long, long long>, long long> b[100005];
map<long long, long long> help;
long long ans[100005];
int main() {
  long long n, i, j, k, l, m;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i].first.second >> a[i].first.first, a[i].second = i;
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    for (j = a[i].first.second; j <= a[i].first.first; j++) {
      if (!help[j]) {
        ans[a[i].second] = j;
        help[j] = 1;
        break;
      }
    }
  }
  for (i = 0; i < n; i++) cout << ans[i] << " ";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
pair<pair<long long, long long>, long long> a[100005];
pair<pair<long long, long long>, long long> b[100005];
map<long long, long long> help;
long long ans[100005];
int main() {
  long long n, i, j, k, l, m;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i].first.second >> a[i].first.first, a[i].second = i;
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    for (j = a[i].first.second; j <= a[i].first.first; j++) {
      if (!help[j]) {
        ans[a[i].second] = j;
        help[j] = 1;
        break;
      }
    }
  }
  for (i = 0; i < n; i++) cout << ans[i] << " ";
}

```