#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e17;

template<typename T, typename E, typename F, typename G, typename H>
class LazySegmentTree {
  int n, height;
  F f; G g; H h; const T ti; const E ei;
  vector<T> dat; vector<E> laz; vector<int> highest;

  void propagate(int k) {
    if(laz[k] != ei) {
      laz[2*k] = h(laz[2*k],laz[k]); laz[2*k+1] = h(laz[2*k+1],laz[k]);
      dat[k] = g(dat[k],laz[k],n/highest[k]);
      laz[k] = ei;
    }
  }
  inline void recalc(int k) {
    while(k >>= 1) dat[k] = f(g(dat[2*k],laz[2*k],n/highest[2*k]),g(dat[2*k+1],laz[2*k+1],n/highest[2*k+1]));
  }
  inline void thrust(int k) {
    for(int i = height; i > 0; i--) propagate(k >> i);
  }
public:
  LazySegmentTree(int _n, F f, G g, H h, T ti, E ei): f(f), g(g), h(h), ti(ti), ei(ei) {
    n = 1; height = 0;
    while(n < _n) n <<= 1, height++;
    dat.assign(2*n,ti); laz.assign(2*n,ei); highest.assign(2*n,0);
    int c = 1;
    for(int i = 1; i < 2*n; i++) {
      if(c*2 <= i) c <<= 1;
      highest[i] = c;
    }
  }
  void set(int k, T x) {
    dat[k+n] = x;
  }
  void build() {
    for(int i = n-1; i >= 0; i--) {
      dat[i] = f(dat[2*i],dat[2*i+1]);
    }
  }
  void update(int a, int b, const E x) {
    thrust(a += n); thrust(b += n-1);
    for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
      if(l & 1) laz[l] = h(laz[l],x), ++l;
      if(r & 1) --r, laz[r] = h(laz[r],x);
    }
    recalc(a); recalc(b);
  }
  T query(int a, int b) {
    thrust(a += n); thrust(b += n-1);
    T l = ti, r = ti; b++;
    while(a < b) {
      if(a & 1) l = f(l,g(dat[a],laz[a],n/highest[a])), a++;
      if(b & 1) b--, r = f(g(dat[b],laz[b],n/highest[b]),r);
      a >>= 1; b >>= 1;
    }
    return f(l,r);
  }
};

int main(){
  int n;
  cin >> n;
  ll x[n], s[n];
  rep(i,n) cin >> x[i] >> s[i];
  auto f = [](ll a, ll b){return max(a,b);};
  auto g = [](ll a, ll b, int l){return a+b;};
  auto h = [](ll a, ll b){return a+b;};
  LazySegmentTree<ll,ll,decltype(f),decltype(g),decltype(h)> lst(n+1,f,g,h,-INF,0);
  rep(i,n) lst.set(i,0);
  lst.build();
  ll ans = -INF;
  rep(i,n) {
    lst.update(i,i+1,x[i]);
    ans = max(ans,lst.query(0,i+1)+s[i]-x[i]);
    lst.update(0,i+1,s[i]);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

In a long narrow forest stretching east-west, there are N beasts. Below, we will call the point that is p meters from the west end Point p. The i-th beast from the west (1 ≤ i ≤ N) is at Point x_i, and can be sold for s_i yen (the currency of Japan) if captured.

You will choose two integers L and R (L ≤ R), and throw a net to cover the range from Point L to Point R including both ends, [L, R]. Then, all the beasts in the range will be captured. However, the net costs R - L yen and your profit will be (the sum of s_i over all captured beasts i) - (R - L) yen.

What is the maximum profit that can be earned by throwing a net only once?

Constraints

* 1 ≤ N ≤ 2 × 10^5
* 1 ≤ x_1 < x_2 < ... < x_N ≤ 10^{15}
* 1 ≤ s_i ≤ 10^9
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
x_1 s_1
x_2 s_2
:
x_N s_N


Output

When the maximum profit is X yen, print the value of X.

Examples

Input

5
10 20
40 50
60 30
70 40
90 10


Output

90


Input

5
10 2
40 5
60 3
70 4
90 1


Output

5


Input

4
1 100
3 200
999999999999999 150
1000000000000000 150


Output

299

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e17;

template<typename T, typename E, typename F, typename G, typename H>
class LazySegmentTree {
  int n, height;
  F f; G g; H h; const T ti; const E ei;
  vector<T> dat; vector<E> laz; vector<int> highest;

  void propagate(int k) {
    if(laz[k] != ei) {
      laz[2*k] = h(laz[2*k],laz[k]); laz[2*k+1] = h(laz[2*k+1],laz[k]);
      dat[k] = g(dat[k],laz[k],n/highest[k]);
      laz[k] = ei;
    }
  }
  inline void recalc(int k) {
    while(k >>= 1) dat[k] = f(g(dat[2*k],laz[2*k],n/highest[2*k]),g(dat[2*k+1],laz[2*k+1],n/highest[2*k+1]));
  }
  inline void thrust(int k) {
    for(int i = height; i > 0; i--) propagate(k >> i);
  }
public:
  LazySegmentTree(int _n, F f, G g, H h, T ti, E ei): f(f), g(g), h(h), ti(ti), ei(ei) {
    n = 1; height = 0;
    while(n < _n) n <<= 1, height++;
    dat.assign(2*n,ti); laz.assign(2*n,ei); highest.assign(2*n,0);
    int c = 1;
    for(int i = 1; i < 2*n; i++) {
      if(c*2 <= i) c <<= 1;
      highest[i] = c;
    }
  }
  void set(int k, T x) {
    dat[k+n] = x;
  }
  void build() {
    for(int i = n-1; i >= 0; i--) {
      dat[i] = f(dat[2*i],dat[2*i+1]);
    }
  }
  void update(int a, int b, const E x) {
    thrust(a += n); thrust(b += n-1);
    for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
      if(l & 1) laz[l] = h(laz[l],x), ++l;
      if(r & 1) --r, laz[r] = h(laz[r],x);
    }
    recalc(a); recalc(b);
  }
  T query(int a, int b) {
    thrust(a += n); thrust(b += n-1);
    T l = ti, r = ti; b++;
    while(a < b) {
      if(a & 1) l = f(l,g(dat[a],laz[a],n/highest[a])), a++;
      if(b & 1) b--, r = f(g(dat[b],laz[b],n/highest[b]),r);
      a >>= 1; b >>= 1;
    }
    return f(l,r);
  }
};

int main(){
  int n;
  cin >> n;
  ll x[n], s[n];
  rep(i,n) cin >> x[i] >> s[i];
  auto f = [](ll a, ll b){return max(a,b);};
  auto g = [](ll a, ll b, int l){return a+b;};
  auto h = [](ll a, ll b){return a+b;};
  LazySegmentTree<ll,ll,decltype(f),decltype(g),decltype(h)> lst(n+1,f,g,h,-INF,0);
  rep(i,n) lst.set(i,0);
  lst.build();
  ll ans = -INF;
  rep(i,n) {
    lst.update(i,i+1,x[i]);
    ans = max(ans,lst.query(0,i+1)+s[i]-x[i]);
    lst.update(0,i+1,s[i]);
  }
  cout << ans << endl;
  return 0;
}

```