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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    // それぞれの地点からスタート、mapで（現在位置、スタート）を保持
    // multisetで横移動の長さを保持
    int H, W;
    cin >> H >> W;

    map<int, int> mp;
    rep(i, W) mp[i] = i;
    multiset<int> vals;
    rep(i, W) vals.insert(0);

    rep(i, H) {
        int l, r;
        scanf("%d%d", &l, &r);
        --l;
        auto it = mp.lower_bound(l);
        int edge = -1; // lとrの中でスタートの右端
        while (it != mp.end() && it->first <= r) {
            edge = max(edge, it->second);
            int x = it->first - it->second;
            mp.erase(it++);
            vals.erase(vals.find(x));
        }
        if (edge >= 0 && r < W) {
            vals.insert(r-edge);
            mp[r] = edge;
        }
        if (vals.empty()) puts("-1");
        else {
            int ans = *vals.begin() + i + 1;
            printf("%d\n", ans);
        }        
    }
    return 0;
}