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 pii pair<int, int>
using namespace std;
const int MAXN = 2e5 + 7;
int A[MAXN], B[MAXN];
set<pii>pr;
multiset<int>ans;
void add(pii x) {
    if (pr.find(x) != pr.end()) return;
    pr.insert(x);
    ans.insert(x.first - x.second);
}
void del(pii x) {
    assert(pr.find(x) != pr.end());
    pr.erase(x);
    ans.erase(ans.find(x.first - x.second));
}
int main() {
    int h, w; scanf("%d %d", &h, &w);
    for (int i = 1; i <= h; i++) scanf("%d %d", &A[i], &B[i]);
    for (int i = 1; i <= w; i++) add({i, i});
    for (int i = 1; i <= h; i++) {
        auto it = pr.lower_bound({A[i], 0});
        int rightmost = -1;
        vector<pii>vec;
        while (it != pr.end() && it->first <= B[i]) {
            rightmost = max(rightmost, it->second);
            vec.push_back(*it);
            it++;
        }
        for (pii &x : vec) del(x);
        if (rightmost != -1 && B[i] != w) {
            add({B[i] + 1, rightmost});
        }
        ans.size() ? printf("%d\n", i + *ans.begin()) : puts("-1");
    }
    return 0;
}