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:
/**
 *   @FileName	f.cpp
 *   @Author	kanpurin
 *   @Created	2020.08.31 16:46:32
**/

#include "bits/stdc++.h" 
using namespace std; 
typedef long long ll;

int main() {
    int h,w;cin >> h >> w;
    set<pair<int,int>> st1; // (現在地,スタート位置)
    multiset<int> st2; // 移動回数
    for (int i = 0; i < w; i++) {
        st1.insert({i,i});
        st2.insert(0);
    }
    for (int i = 0; i < h; i++) {
        int a,b;cin >> a >> b;
        // 現在地が[a,b+1]にあるものをb+1に移動させる.
        // b=wなら∞に移動.
        // 移動させるものの中で一番右のものだけ残す.
        a--;b--;
        auto l = st1.lower_bound({a,-1});
        auto r = st1.upper_bound({b+1,w});
        if (l != r) { // ある
            r--;
            while(l != r) {
                st2.erase(st2.find(l->first-l->second));
                l = st1.erase(l);
            }
            st2.erase(st2.find(r->first-r->second));
            st1.erase(r);
            // b+1への移動
            if (b+1 != w) {
                st1.insert({b+1,r->second});
                st2.insert(b+1-r->second);
            }
        }
        if (st1.empty()) cout << -1 << endl;
        else cout << *st2.begin() + i + 1 << endl;
    }
    return 0;
}