Problem: All modern mobile applications are divided into free and paid. Even a single application developers often release two versions: a paid version without ads and a free version with ads.

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Suppose that a paid version of the app costs p (p is an integer) rubles, and the free version of the application contains c ad banners. Each user can be described by two integers: ai — the number of rubles this user is willing to pay for the paid version of the application, and bi — the number of banners he is willing to tolerate in the free version.

The behavior of each member shall be considered strictly deterministic:

  * if for user i, value bi is at least c, then he uses the free version, 
  * otherwise, if value ai is at least p, then he buys the paid version without advertising, 
  * otherwise the user simply does not use the application. 



Each user of the free version brings the profit of c × w rubles. Each user of the paid version brings the profit of p rubles.

Your task is to help the application developers to select the optimal parameters p and c. Namely, knowing all the characteristics of users, for each value of c from 0 to (max bi) + 1 you need to determine the maximum profit from the application and the corresponding parameter p.

Input

The first line contains two integers n and w (1 ≤ n ≤ 105; 1 ≤ w ≤ 105) — the number of users and the profit from a single banner. Each of the next n lines contains two integers ai and bi (0 ≤ ai, bi ≤ 105) — the characteristics of the i-th user.

Output

Print (max bi) + 2 lines, in the i-th line print two integers: pay — the maximum gained profit at c = i - 1, p (0 ≤ p ≤ 109) — the corresponding optimal app cost. If there are multiple optimal solutions, print any of them.

Examples

Input

2 1
2 0
0 2


Output

0 3
3 2
4 2
2 2


Input

3 1
3 1
2 2
1 3


Output

0 4
3 4
7 3
7 2
4 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O2,unroll-loops")
using namespace std;
const int inf = 1000000010;
const long long INF = 1000000000000001000LL;
const int mod = 1000000007;
const int MAXN = 100200, SQ = 320, NSQ = MAXN / SQ + 1;
int n, m, w, u, v, x, y, t, a, b, maxb;
int A[MAXN], B[MAXN], ans[MAXN], lazy[NSQ];
pair<long long, long long> X[MAXN];
vector<int> vec[MAXN];
struct CHT {
  pair<long long, int> cht[SQ];
  int L, R;
  inline long long Inter(pair<long long, long long> x,
                         pair<long long, long long> y) {
    if (x.first == y.first) return (x <= y ? -INF : INF);
    return (x.second - y.second) / (y.first - x.first) +
           ((x.second - y.second) % (y.first - x.first) > 0);
  }
  inline void Add(int a) {
    while (R && Inter(X[cht[R - 1].second], X[a]) <= cht[R - 1].first) R--;
    if (!R)
      cht[R++] = {-INF, a};
    else
      cht[R] = {Inter(X[cht[R - 1].second], X[a]), a}, R++;
  }
  inline pair<long long, long long> Get(long long x) {
    while (L + 1 < R && cht[L + 1].first <= x) L++;
    int i = cht[L].second;
    return {X[i].first * x + X[i].second, i};
  }
} cht[NSQ];
void Build(int id) {
  int tl = id * SQ, tr = min(MAXN, tl + SQ);
  cht[id].L = cht[id].R = 0;
  for (int i = tl; i < tr; i++) {
    X[i].second += lazy[id] * X[i].first;
    cht[id].Add(i);
  }
  lazy[id] = 0;
}
void Add(int l, int r, int val) {
  int idl = -1, idr = -1;
  while (l < r && l % SQ) idl = l / SQ, X[l].second += val * l, l++;
  while (l < r && r % SQ) idr = r / SQ, r--, X[r].second += val * r;
  while (l < r) lazy[l / SQ] += val, l += SQ;
  if (idl != -1) Build(idl);
  if (idr != -1) Build(idr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> w;
  for (int i = 1; i <= n; i++) {
    cin >> A[i] >> B[i];
    vec[B[i]].push_back(i);
    maxb = max(maxb, B[i]);
  }
  for (int i = 0; i < MAXN; i++) X[i].first = i;
  for (int i = 0; i * SQ < MAXN; i++) Build(i);
  int ted = n;
  for (int i = 0; i <= maxb + 1; i++) {
    pair<long long, long long> p = {-inf, -1};
    for (int j = 0; j * SQ + SQ <= MAXN; j++) p = max(p, cht[j].Get(lazy[j]));
    cout << p.first + 1ll * ted * i * w << " " << p.second << "\n";
    for (int j : vec[i]) Add(0, A[j] + 1, +1), ted--;
  }
  return 0;
}