#include <bits/stdc++.h>
using namespace std;
using Int = long long;
struct SegTreeMax {
    vector<Int> A;
    int N;
    SegTreeMax(int n) {
        N = 1;
        while (N < n) N *= 2;
        A.resize(2*N, LONG_LONG_MIN);
    }
    void set(int i, Int val) {
        i += N;
        A[i] = val;
        i /= 2;
        while (i > 0) {
            A[i] = max(A[2*i+0], A[2*i+1]);
            i /= 2;
        }
    }
    Int query(int L, int R) {
        L += N, R += N;
        Int ans = LONG_LONG_MIN;
        while (L < R) {
            if (R & 1) ans = max(ans, A[--R]);
            if (L & 1) ans = max(ans, A[L++]);
            L /= 2, R /= 2;
        }
        return ans;
    }
};
int main()
{
    int N; cin >> N;
    vector<Int> X(N), S(N);
    for (int i = 0; i < N; i++) cin >> X[i] >> S[i];
    vector<Int> sum(N+1);
    for (int i = 0; i < N; i++) sum[i+1] = sum[i] + S[i];
    SegTreeMax st(N+1);
    st.set(0, 0);
    for (int i = 0; i < N; i++) st.set(i+1, sum[i+1] - X[i]);
    Int ans = 0;
    for (int i = 0; i < N; i++) {
        Int score = st.query(i+1, N+1) - st.query(i, i+1) + X[i] - (i > 0 ? X[i-1] : 0);
        ans = max(ans, score);
    }
    cout << ans << endl;
    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
using Int = long long;
struct SegTreeMax {
    vector<Int> A;
    int N;
    SegTreeMax(int n) {
        N = 1;
        while (N < n) N *= 2;
        A.resize(2*N, LONG_LONG_MIN);
    }
    void set(int i, Int val) {
        i += N;
        A[i] = val;
        i /= 2;
        while (i > 0) {
            A[i] = max(A[2*i+0], A[2*i+1]);
            i /= 2;
        }
    }
    Int query(int L, int R) {
        L += N, R += N;
        Int ans = LONG_LONG_MIN;
        while (L < R) {
            if (R & 1) ans = max(ans, A[--R]);
            if (L & 1) ans = max(ans, A[L++]);
            L /= 2, R /= 2;
        }
        return ans;
    }
};
int main()
{
    int N; cin >> N;
    vector<Int> X(N), S(N);
    for (int i = 0; i < N; i++) cin >> X[i] >> S[i];
    vector<Int> sum(N+1);
    for (int i = 0; i < N; i++) sum[i+1] = sum[i] + S[i];
    SegTreeMax st(N+1);
    st.set(0, 0);
    for (int i = 0; i < N; i++) st.set(i+1, sum[i+1] - X[i]);
    Int ans = 0;
    for (int i = 0; i < N; i++) {
        Int score = st.query(i+1, N+1) - st.query(i, i+1) + X[i] - (i > 0 ? X[i-1] : 0);
        ans = max(ans, score);
    }
    cout << ans << endl;
    return 0;
}
```