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;
#define int long long
typedef pair<int,int> P;
int INF = 1e18;
int mod = 1e9+7;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
signed main() {
    int H,W;
    cin >> H >> W;
    map<int,int>Mp;
    multiset<int>st;
    for(int i = 0; i < W; i++) {
        Mp[i] = i;
        st.insert(0);
    }
    for(int i = 0; i < H; i++) {
        int A,B;
        cin >> A >> B;
        A--;B--;
        auto it = Mp.lower_bound(A);
        int mx = -1;
        while (it != Mp.end() && it->first <= B+1) {
            st.erase(st.find(it->first-it->second));
            mx = max(mx,it->second);
            Mp.erase(it++);
        }
        if(B+1 != W && mx != -1) {
            Mp[B+1] = mx;
            st.insert(B+1-Mp[B+1]);
        }
        if(st.size() == 0) {
            cout << -1 << endl;
        }
        else {
            cout << *st.begin()+i+1 << endl;
        }
    }
}