Problem: There is a grid of squares with H+1 horizontal rows and W vertical columns.

You will start at one of the squares in the top row and repeat moving one square right or down. However, for each integer i from 1 through H, you cannot move down from the A_i-th, (A_i + 1)-th, \ldots, B_i-th squares from the left in the i-th row from the top.

For each integer k from 1 through H, find the minimum number of moves needed to reach one of the squares in the (k+1)-th row from the top. (The starting square can be chosen individually for each case.) If, starting from any square in the top row, none of the squares in the (k+1)-th row can be reached, print `-1` instead.

Constraints

* 1 \leq H,W \leq 2\times 10^5
* 1 \leq A_i \leq B_i \leq W

Input

Input is given from Standard Input in the following format:


H W
A_1 B_1
A_2 B_2
:
A_H B_H


Output

Print H lines. The i-th line should contain the answer for the case k=i.

Example

Input

4 4
2 4
1 1
2 3
2 4


Output

1
3
6
-1

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>

int main() {
  constexpr int kInfinity = 1000000;
  int H, W;
  std::cin >> H >> W;
  std::vector<int> A(H), B(H);
  for (int i = 0; i < H; ++i) {
    std::cin >> A[i] >> B[i];
    --A[i];
  }

  std::map<int, int> costs;
  for (int i = 0; i < W; ++i) {
    costs[i] = 0;
  }
  std::map<int, int> counts;
  counts[0] = W;
  for (int i = 0; i < H; ++i) {
    if (counts.empty()) {
      std::cout << -1 << std::endl;
      continue;
    }
    const auto begin = costs.lower_bound(A[i]);
    const auto end = costs.upper_bound(B[i]);
    int best_cost = kInfinity;
    for (auto iter = begin; iter != end; ++iter) {
      const auto [x, cost] = *iter;
      best_cost = cost + B[i] - x;
      if (--counts[cost] == 0) {
        counts.erase(cost);
      }
    }
    costs.erase(begin, end);
    if (best_cost < kInfinity && B[i] < W) {
      ++counts[best_cost];
      costs[B[i]] = best_cost;
    }
    if (counts.empty()) {
      std::cout << -1 << std::endl;
    } else {
      std::cout << (counts.begin()->first + i + 1) << std::endl;
    }
  }
  return 0;
}