// ※※※ 解答不能 ※※※
// tourist氏.
// https://atcoder.jp/contests/cf17-relay-open/submissions/8927539
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define repx(i, a, b) for(int i = a; i < b; i++)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)

int main(){
    int n;
    scanf("%d", &n);
    vector<LL> x(n);
    vector<int> s(n);
    rep(i, n) scanf("%lld %d", &x[i], &s[i]);
    LL ans = 0, best = (LL) -9e18, sum = 0;
    rep(i, n){
        best = max(best, -sum + x[i]);
        sum += s[i];
        ans = max(ans, sum - x[i] + best);
    }
    printf("%lld\n", ans);
    return 0;
}
### Prompt

Please formulate a cpp solution to 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
// ※※※ 解答不能 ※※※
// tourist氏.
// https://atcoder.jp/contests/cf17-relay-open/submissions/8927539
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define repx(i, a, b) for(int i = a; i < b; i++)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)

int main(){
    int n;
    scanf("%d", &n);
    vector<LL> x(n);
    vector<int> s(n);
    rep(i, n) scanf("%lld %d", &x[i], &s[i]);
    LL ans = 0, best = (LL) -9e18, sum = 0;
    rep(i, n){
        best = max(best, -sum + x[i]);
        sum += s[i];
        ans = max(ans, sum - x[i] + best);
    }
    printf("%lld\n", ans);
    return 0;
}
```