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;
template <typename T>
bool chmax(T& a, const T& b) {
  return a < b ? (a = b, true) : false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  return b < a ? (a = b, true) : false;
}
int N;
const int MAX_N = 100000;
int T[MAX_N];
long long L[MAX_N], R[MAX_N];
vector<long long> VP;
unsigned long long dat[MAX_N * 3 / 64];
int main() {
  scanf("%d", &N);
  for (int i = (int)(0); i < (int)(N); i++) {
    scanf("%d%lld%lld", &T[i], &L[i], &R[i]);
    VP.push_back(L[i]);
    VP.push_back(R[i] + 1);
  }
  sort(VP.begin(), VP.end());
  VP.erase(unique(VP.begin(), VP.end()), VP.end());
  if (VP.front() != 1) {
    for (int i = (int)(0); i < (int)(N); i++) {
      puts("1");
    }
    return 0;
  }
  for (int i = (int)(0); i < (int)(N); i++) {
    int l = lower_bound(VP.begin(), VP.end(), L[i]) - VP.begin();
    int r = lower_bound(VP.begin(), VP.end(), R[i] + 1) - VP.begin();
    while ((l & 63) && l < r) {
      int x = l / 64, y = l % 64;
      if (T[i] == 1) {
        dat[x] |= 1llu << y;
      } else if (T[i] == 2) {
        dat[x] &= ~(1llu << y);
      } else {
        dat[x] ^= 1llu << y;
      }
      l++;
    }
    while ((r & 63) && l < r) {
      r--;
      int x = r / 64, y = r % 64;
      if (T[i] == 1) {
        dat[x] |= 1llu << y;
      } else if (T[i] == 2) {
        dat[x] &= ~(1llu << y);
      } else {
        dat[x] ^= 1llu << y;
      }
    }
    while (l < r) {
      if (T[i] == 1) {
        dat[l / 64] = ~0llu;
      } else if (T[i] == 2) {
        dat[l / 64] = 0;
      } else {
        dat[l / 64] ^= ~0llu;
      }
      l += 64;
    }
    for (int j = (int)(0); j < (int)(N); j++)
      if (dat[j] != ~0llu) {
        for (int k = (int)(0); k < (int)(64); k++)
          if ((dat[j] >> k & 1) == 0) {
            printf("%lld\n", VP[j * 64 + k]);
            break;
          }
        break;
      }
  }
  return 0;
}