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
#include <bits/stdc++.h>
using namespace std;
const int N = 105010;
const int M = N / 300;
struct Data {
  int a, b;
  Data(int a = 0, int b = 0) : a(a), b(b) {}
  bool operator<(const Data &O) const { return b < O.b; }
};
int n, w;
int que[M][310], l[M], r[M];
long long sum[M][310], cnt[M];
pair<long long, int> Insert(int x) {
  int id = x / 300;
  pair<long long, int> ans = make_pair(0, n + 1);
  for (int i = 0; i < id; i++) {
    cnt[i]++;
    while (l[i] + 1 < r[i]) {
      long long a = sum[i][que[i][l[i]]] + cnt[i];
      int ia = que[i][l[i]] + i * 300;
      long long b = sum[i][que[i][l[i] + 1]] + cnt[i];
      int ib = que[i][l[i] + 1] + i * 300;
      if (ia * a <= ib * b)
        l[i]++;
      else
        break;
    }
    long long a = sum[i][que[i][l[i]]] + cnt[i];
    int ia = que[i][l[i]] + i * 300;
    ans = max(ans, make_pair(a * ia, ia));
  }
  int y = x / 300 * 300;
  for (int i = 0; i <= x % 300; i++) sum[id][i]++;
  l[id] = r[id] = 0;
  for (int i = 0; i < 300; i++) {
    while (r[id] > l[id]) {
      long long a = sum[id][que[id][r[id] - 1]] + cnt[id];
      int ia = que[id][r[id] - 1] + y;
      long long b = sum[id][i] + cnt[id];
      int ib = i + y;
      if (ia * a <= ib * b)
        r[id]--;
      else
        break;
    }
    while (r[id] > l[id] + 1) {
      long long a = sum[id][que[id][r[id] - 2]] + cnt[id];
      int ia = que[id][r[id] - 2] + y;
      long long b = sum[id][que[id][r[id] - 1]] + cnt[id];
      int ib = que[id][r[id] - 1] + y;
      long long c = sum[id][i] + cnt[id];
      int ic = i + y;
      int A = (ib - ia);
      long long B = (ia * a - ib * b + A - 1) / A;
      if (ic * c - ib * b + B * (ic - ib) >= 0)
        r[id]--;
      else
        break;
    }
    que[id][r[id]++] = i;
  }
  long long b = sum[id][que[id][l[id]]] + cnt[id];
  int ib = que[id][l[id]] + y;
  ans = max(ans, make_pair(b * ib, ib));
  return ans;
}
int main() {
  scanf("%d%d", &n, &w);
  vector<Data> vt;
  int mx = 0;
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    vt.push_back(Data(a, b));
    mx = max(mx, b + 1);
  }
  for (int i = 0; i < M; i++) {
    que[i][r[i]++] = 299;
  }
  sort(vt.begin(), vt.end());
  pair<long long, int> ans = make_pair(0, n + 1);
  for (int i = 0, j = 0; i <= mx; i++) {
    while (j < vt.size() && vt[j].b < i) {
      ans = max(ans, Insert(vt[j].a));
      j++;
    }
    cout << ans.first + 1ll * i * w * (n - j) << " " << ans.second << endl;
  }
  return 0;
}
