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>

const int N = 200010;

struct Edge{
    int start, end;

    bool operator <(const Edge &e)const{
        return end < e.end;
    }
};

int a[N], b[N];

int main(){
    int h, w;
    scanf("%d%d", &h, &w);
    for (int i = 0; i < h; ++ i){
        scanf("%d%d", &a[i], &b[i]);
    }
    std::set <Edge> set;
    for (int i = 1; i <= w; ++ i){
        set.insert(Edge{i, i});
    }
    std::multiset <int> answer;
    for (int i = 1; i <= w; ++ i){
        answer.insert(0);
    }
    for (int i = 0; i < h; ++ i){
        auto u = set.lower_bound(Edge{0, a[i]});
        int max = 0;
        while (u != set.end() && u -> end <= b[i] + 1){
            auto v = answer.find(u -> end - u -> start);
            answer.erase(v);
            max = std::max(max, u -> start);
            u = set.erase(u);
        }
        if (max > 0 && b[i] + 1 <= w){
            set.insert(Edge{max, b[i] + 1});
            answer.insert(b[i] + 1 - max);
        }
        if (answer.empty()){
            puts("-1");
            continue;
        }
        printf("%d\n", *(answer.begin()) + i + 1);
    }
    return 0;
}