Problem: You are given a set of integer numbers, initially it is empty. You should perform n queries.

There are three different types of queries: 

  * 1 l r — Add all missing numbers from the interval [l, r]
  * 2 l r — Remove all present numbers from the interval [l, r]
  * 3 l r — Invert the interval [l, r] — add all missing and remove all present numbers from the interval [l, r]



After each query you should output MEX of the set — the smallest positive (MEX ≥ 1) integer number which is not presented in the set.

Input

The first line contains one integer number n (1 ≤ n ≤ 105).

Next n lines contain three integer numbers t, l, r (1 ≤ t ≤ 3, 1 ≤ l ≤ r ≤ 1018) — type of the query, left and right bounds.

Output

Print MEX of the set after each query.

Examples

Input

3
1 3 4
3 1 6
2 1 3


Output

1
3
1


Input

4
1 1 3
3 5 6
2 4 4
3 1 6


Output

4
4
4
1

Note

Here are contents of the set after each query in the first example:

  1. {3, 4} — the interval [3, 4] is added 
  2. {1, 2, 5, 6} — numbers {3, 4} from the interval [1, 6] got deleted and all the others are added 
  3. {5, 6} — numbers {1, 2} got deleted 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 1e5;
struct Node {
  long long l, r;
  mutable bool v;
  Node(long long _l, long long _r = 0, bool _v = 0) : l(_l), r(_r), v(_v) {}
};
inline bool operator<(const Node& x, const Node& y) { return x.l < y.l; }
std::set<Node> s;
long long n;
inline void read(long long& x) {
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
}
inline std::set<Node>::iterator split(long long pos) {
  std::set<Node>::iterator it = s.lower_bound(Node(pos));
  if (it->l == pos && it != s.end()) return it;
  it--;
  long long l = it->l, r = it->r;
  bool v = it->v;
  s.erase(it);
  s.insert(Node(l, pos - 1, v));
  return s.insert(Node(pos, r, v)).first;
}
inline void assign(long long l, long long r, bool v) {
  std::set<Node>::iterator itr = split(r + 1), itl = split(l);
  s.erase(itl, itr);
  s.insert(Node(l, r, v));
}
inline void reverse(long long l, long long r) {
  std::set<Node>::iterator itr = split(r + 1), itl = split(l);
  for (std::set<Node>::iterator it = itl; it != itr; it++) it->v ^= 1;
}
inline long long query() {
  for (std::set<Node>::iterator it = s.begin(); it != s.end(); it++)
    if (it->v == 0) return it->l;
  return -1;
}
int main() {
  read(n);
  s.insert(Node(1, 1e18, 0));
  while (n--) {
    long long opt, l, r;
    read(opt), read(l), read(r);
    if (opt == 1) assign(l, r, 1);
    if (opt == 2) assign(l, r, 0);
    if (opt == 3) reverse(l, r);
    printf("%lld\n", query());
  }
  return 0;
}