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;
const int INF = 0x3fffffff;

struct SegmentTree{
    int n;
    vector<int> data;
    SegmentTree(int n): n(n), data(n * 2){}
    void set(int i, int x){
        i += n;
        data[i] = x;
        while(i /= 2) data[i] = min(data[i * 2], data[i * 2 + 1]);
    }
    int get(){
        return data[1];
    }
    int operator[](int i){ return data[n + i]; }
};
int main(){
    int h, w;
    cin >> h >> w;
    SegmentTree seg(w);
    map<int, int> s;
    for(int i = 0; i < w; i++) s[i] = i;
    for(int i = 1; i <= h; i++){
        int a, b;
        cin >> a >> b;
        a--;
        auto p = s.lower_bound(a);
        int mx = -1;
        while(p != s.end()){
            const auto [at, start] = *p;
            if(at >= b) break;
            mx = start;
            seg.set(at, INF);
            p = s.erase(p);
        }
        if(mx != -1 && b != w && seg[b] == INF){
            s[b] = mx;
            seg.set(b, b - mx);
        }
        if(seg.get() == INF) puts("-1");
        else cout << seg.get() + i << '\n';
    }
}