Vasya got really tired of these credits (from problem F) and now wants to earn the money himself! He decided to make a contest to gain a profit.

Vasya has n problems to choose from. They are numbered from 1 to n. The difficulty of the i-th problem is d_i. Moreover, the problems are given in the increasing order by their difficulties. The difficulties of all tasks are pairwise distinct. In order to add the i-th problem to the contest you need to pay c_i burles to its author. For each problem in the contest Vasya gets a burles.

In order to create a contest he needs to choose a consecutive subsegment of tasks.

So the total earnings for the contest are calculated as follows: 

  * if Vasya takes problem i to the contest, he needs to pay c_i to its author; 
  * for each problem in the contest Vasya gets a burles; 
  * let gap(l, r) = max_{l ≤ i < r} (d_{i + 1} - d_i)^2. If Vasya takes all the tasks with indices from l to r to the contest, he also needs to pay gap(l, r). If l = r then gap(l, r) = 0. 



Calculate the maximum profit that Vasya can earn by taking a consecutive segment of tasks.

Input

The first line contains two integers n and a (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ a ≤ 10^9) — the number of proposed tasks and the profit for a single problem, respectively.

Each of the next n lines contains two integers d_i and c_i (1 ≤ d_i, c_i ≤ 10^9, d_i < d_{i+1}).

Output

Print one integer — maximum amount of burles Vasya can earn.

Examples

Input


5 10
1 15
5 3
6 11
7 2
11 22


Output


13


Input


3 5
1 8
2 19
3 11


Output


0
#include <bits/stdc++.h>
using namespace std;
struct DSU {
  struct Node {
    int pr;
    long long max_sum;
    long long max_pref;
    long long sum;
    long long max_suff;
  };
  vector<Node> t;
  multiset<long long> mx;
  DSU(int n, vector<int>& c) {
    t.resize(n);
    for (int i = 0; i < n; ++i) {
      t[i].pr = i;
      t[i].sum = c[i];
      t[i].max_sum = max(0, c[i]);
      t[i].max_pref = max(0, c[i]);
      t[i].max_suff = max(0, c[i]);
      mx.insert(t[i].max_sum);
    }
  }
  int f(int v) {
    if (t[v].pr != v) {
      t[v].pr = f(t[v].pr);
    }
    return t[v].pr;
  }
  void merge(int x, int y) {
    x = f(x);
    y = f(y);
    if (x == y) {
      return;
    }
    t[x].pr = y;
    mx.erase(mx.lower_bound(t[x].max_sum));
    mx.erase(mx.lower_bound(t[y].max_sum));
    Node new_node;
    new_node.sum = t[x].sum + t[y].sum;
    new_node.pr = y;
    new_node.max_sum = max(t[y].max_sum, t[x].max_sum);
    new_node.max_sum = max(new_node.max_sum, t[x].max_suff + t[y].max_pref);
    new_node.max_pref = max(t[x].max_pref, t[x].sum + t[y].max_pref);
    new_node.max_suff = max(t[y].max_suff, t[y].sum + t[x].max_suff);
    new_node.max_sum = max(new_node.max_sum, 0LL);
    new_node.max_pref = max(new_node.max_pref, 0LL);
    new_node.max_suff = max(new_node.max_suff, 0LL);
    t[y] = new_node;
    mx.insert(t[y].max_sum);
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  long long m;
  cin >> n >> m;
  vector<int> d(n), c(n);
  for (int i = 0; i < n; ++i) {
    cin >> d[i] >> c[i];
    c[i] = m - c[i];
  }
  set<pair<int, int> > q;
  for (int i = 0; i + 1 < n; ++i) {
    q.insert({d[i + 1] - d[i], i});
  }
  q.insert({-1, -1});
  DSU t(n, c);
  long long ans = 0;
  while (q.size()) {
    auto i = *q.begin();
    q.erase(q.begin());
    if (i.second != -1) {
      t.merge(i.second, i.second + 1);
      ans = max(ans, *--t.mx.end() - i.first * 1LL * i.first);
    } else {
      ans = max(ans, *--t.mx.end());
    }
  }
  cout << ans << "\n";
}
