Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4 4
2 4
1 1
2 3
2 4

Sample Output 1

1
3
6
-1

Let (i,j) denote the square at the i-th row from the top and j-th column from the left.

For k=1, we need one move such as (1,1) → (2,1).

For k=2, we need three moves such as (1,1) → (2,1) → (2,2) → (3,2).

For k=3, we need six moves such as (1,1) → (2,1) → (2,2) → (3,2) → (3,3) → (3,4) → (4,4).

For k=4, it is impossible to reach any square in the fifth row from the top.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}

#ifdef DEBUG
void debug() { cerr << "\n"; }
template <class T> void debug(const T &x) { cerr << x << "\n"; }
template <class T, class... Args> void debug(const T &x, const Args &... args) {
    cerr << x << " ";
    debug(args...);
}
template <class T> void debugVector(const vector<T> &v) {
    for(const T &x : v) {
        cerr << x << " ";
    }
    cerr << "\n";
}
#else
template <class T, class... Args>
void debug(const T &x, const Args &... args) {}
template <class T> void debugVector(const vector<T> &v) {}
#endif

using ll = long long;
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
const double EPS = 1e-7;
const int INF = 1 << 30;
const ll LLINF = 1LL << 60;
constexpr int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int H, W;
    cin >> H >> W;
    map<int, int> start;
    multiset<int> vals;
    for(int i = 0; i < W; i++) {
        start[i] = i;
        vals.insert(0);
    }
    for(int i = 0; i < H; i++) {
        int l, r;
        cin >> l >> r;
        l--;
        auto it = start.lower_bound(l);
        int mxr = -1;
        while(it != start.end() && it->first <= r) {
            chmax(mxr, it->second);
            vals.erase(vals.find(it->first - it->second));
            start.erase(it++);
        }
        if(mxr != -1 && r < W) {
            vals.insert(r - mxr);
            start[r] = mxr;
        }
        int ans = -1;
        if(!vals.empty()) {
            ans = *vals.begin();
            ans += i + 1;
        }
        cout << ans << "\n";
    }
}