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>
using namespace std;
const int maxn = 1e5 + 5;
const int MAGIC = 400;
int n, lima, limb, w;
long long coef[maxn];
struct point {
  long long a, b;
  point(){};
  point(long long a, long long b) : a(a), b(b){};
  point operator-(point other) { return point(a - other.a, b - other.b); }
  long long operator*(point other) { return a * other.b - b * other.a; };
};
struct CHT {
  int L, R, it, m, x = 0;
  vector<point> hull;
  void add(int pos) {
    for (int i = L; i <= pos; ++i) coef[i] += i;
    build();
  }
  void build() {
    hull.clear();
    m = it = 0;
    for (int i = L; i <= R; ++i) {
      while (m >= 2 &&
             (hull[m - 2] - hull[m - 1]) * (point(i, coef[i]) - hull[m - 1]) <=
                 0)
        --m, hull.pop_back();
      ++m;
      hull.push_back(point(i, coef[i]));
    }
  }
  pair<long long, long long> ask() {
    while (it + 1 < hull.size() &&
           x * hull[it].a + hull[it].b <= x * hull[it + 1].a + hull[it + 1].b)
      ++it;
    return make_pair(x * hull[it].a + hull[it].b, hull[it].a);
  }
} b[MAGIC];
pair<int, int> a[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("A"
            ".INP",
            "r")) {
    freopen(
        "A"
        ".INP",
        "r", stdin);
    freopen(
        "A"
        ".OUT",
        "w", stdout);
  }
  cin >> n >> w;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].second >> a[i].first;
    limb = max(limb, a[i].first + 1);
    lima = max(lima, a[i].second + 1);
  }
  for (int i = 0; i <= lima; i += MAGIC) {
    b[i / MAGIC].L = i;
    b[i / MAGIC].R = min(lima, i + MAGIC - 1);
    b[i / MAGIC].build();
  }
  sort(a + 1, a + n + 1);
  int it = 1;
  for (int i = 0; i <= limb; ++i) {
    auto add = [&](int x) {
      for (int i = 0; i <= x; i += MAGIC) {
        if (i + MAGIC > x) {
          b[i / MAGIC].add(x);
          break;
        }
        b[i / MAGIC].x++;
      }
    };
    while (it <= n && a[it].first < i) add(a[it++].second);
    pair<long long, long long> res = make_pair(-1, 0);
    for (int j = 1; j <= lima; j += MAGIC) {
      res = max(res, b[j / MAGIC].ask());
    }
    cout << res.first + 1ll * w * i * (n - it + 1) << " " << res.second << '\n';
  }
}