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 M = 1e5 + 5, LG = 26, SM = 350, inf = 1e9 + 1, mod = 95542721;
set<int> se;
set<int>::iterator vasat;
int rotbe;
int ted[M];
long long int ans;
int ansp;
int rng = 3000;
void ins(int x) {
  rotbe += (x >= *vasat);
  ted[x]++;
  se.insert(x);
  auto vst2 = vasat;
  int rtbt2 = rotbe;
  if (x >= *vasat) {
    auto ite = vasat;
    int rot = rotbe;
    for (int rep = 0; rep < rng; rep++) {
      if (1LL * (*ite) * rot > ans) {
        rtbt2 = rot;
        vst2 = ite;
        ans = 1LL * (*ite) * rot;
        ansp = *ite;
      }
      rot -= ted[*ite];
      ite++;
      if (ite == se.end()) break;
    }
  } else {
    auto ite = vasat;
    int rot = rotbe;
    for (int rep = 0; rep < rng; rep++) {
      if (1LL * (*ite) * rot > ans) {
        rtbt2 = rot;
        vst2 = ite;
        ans = 1LL * (*ite) * rot;
        ansp = *ite;
      }
      if (ite == se.begin()) break;
      ite--;
      rot += ted[*ite];
    }
  }
  rotbe = rtbt2;
  vasat = vst2;
}
int n, w;
vector<int> b[M];
int main() {
  ted[0]++;
  se.insert(0);
  vasat = se.begin();
  rotbe = 1;
  scanf("%d%d", &n, &w);
  int mxb = 0;
  for (int i = 0; i < n; i++) {
    int ai, bi;
    scanf("%d%d", &ai, &bi);
    mxb = max(mxb, bi);
    b[bi].push_back(ai);
  }
  mxb += 2;
  int tnaf = n;
  for (int c = 0; c < mxb; c++) {
    printf("%lld %d\n", ans + 1LL * tnaf * c * w, ansp);
    for (int y : b[c]) {
      ins(y);
      tnaf--;
    }
  }
}
