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>
using namespace std;
const int N = 1e5 + 50;
struct Node {
  long long l, r;
  mutable long long v;
  Node(long long L = -1, long long R = -1, long long V = 0)
      : l(L), r(R), v(V) {}
  bool operator<(const Node& x) const { return l < x.l; }
};
set<Node> st;
long long n;
inline long long Read() {
  long long x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x;
}
set<Node>::iterator Split(long long pos) {
  set<Node>::iterator it = st.lower_bound(Node(pos));
  if (it != st.end() && it->l == pos) {
    return it;
  }
  it--;
  long long L = it->l, R = it->r, V = it->v;
  st.erase(it);
  st.insert(Node(L, pos - 1, V));
  return st.insert(Node(pos, R, V)).first;
}
void Assign(long long l, long long r, long long x) {
  set<Node>::iterator itr = Split(r + 1);
  set<Node>::iterator itl = Split(l);
  st.erase(itl, itr);
  st.insert(Node(l, r, x));
}
void Change(long long l, long long r) {
  set<Node>::iterator itr = Split(r + 1);
  set<Node>::iterator itl = Split(l);
  for (set<Node>::iterator it = itl; it != itr; it++) it->v = !it->v;
}
long long Query_MEX() {
  for (set<Node>::iterator it = st.begin(); it != st.end(); it++) {
    if (!it->v) return it->l;
  }
}
int main() {
  n = Read();
  st.insert(Node(1, 1e19, 0));
  for (long long i = 1; i <= n; i++) {
    long long t = Read(), l = Read(), r = Read();
    if (t == 1) Assign(l, r, 1);
    if (t == 2) Assign(l, r, 0);
    if (t == 3) Change(l, r);
    printf("%lld\n", Query_MEX());
  }
  return 0;
}