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 <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int H, W;
    cin >> H >> W;
    vector<int> A(H), B(H);
    for (int i = 0; i < H; ++i) {
        cin >> A[i] >> B[i];
    }
    set<pair<int, int>> S; // cur, start
    vector<int> difference_count(W, 0); // cur - start
    for (int i = 1; i <= W; ++i) {
        S.insert({i, i});
    }
    difference_count[0] = W;
    int min_difference = 0;
    for (int i = 0; i < H; ++i) {
        auto L = S.upper_bound({A[i], 0});
        auto R = S.upper_bound({B[i] + 2, 0});
        if (L != S.end() && L != R) {
            int max_start = 0;
            for (auto it = L; it != R; ++it) {
                int cur = it->first, start = it->second;
                --difference_count[cur - start];
                max_start = max(start, max_start);
            }
            S.erase(L, R);
            if (B[i] + 1 <= W) {
                S.insert({B[i] + 1, max_start});
                ++difference_count[B[i] + 1 - max_start];
            }
        }
        if (S.empty()) {
            cout << "-1\n";
        }
        else {
            while (difference_count[min_difference] == 0) {
                ++min_difference;
            }
            cout << i + 1 + min_difference << '\n';
        }
    }
    return 0;
}