#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define il inline
#define outy printf("Yes\n")
#define outn printf("No\n")
#define outf printf("First\n")
#define outs printf("Second\n")
#define FOR(a,b,c) for(int a=b;a<=c;a++)
#define FORD(a,b,c) for(int a=b;a>=c;a--)
#define FORL(a,b) for(int a=head[b];a;a=nxt[a])
#define in(a)  a=read()
#define in2(a,b)  a=read(),b=read()
#define in3(a,b,c)  a=read(),b=read(),c=read()
#define in4(a,b,c,d)  a=read(),b=read(),c=read(),d=read()
#define inl(a)  a=readl()
#define inl2(a,b)  a=readl(),b=readl()
#define inl3(a,b,c)  a=readl(),b=readl(),c=readl()
#define inl4(a,b,c,d)  a=readl(),b=readl(),c=readl(),d=readl()
#define out(a) printf("%d\n",a)
#define out2(a,b) printf("%d %d\n",a,b)
#define out3(a,b,c) printf("%d %d %d\n",a,b,c)
#define out4(a,b,c,d) printf("%d %d %d %d\n",a,b,c,d)
#define outl(a) printf("%lld\n",a)
#define outl2(a,b) printf("%lld %lld\n",a,b)
#define outl3(a,b,c) printf("%lld %lld %lld\n",a,b,c)
#define outl4(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)
#define clr(x,y) memset(x,y,sizeof(x))
inline int read(){
    char c;int s=0,t=1;
    while(!isdigit(c=getchar()))if(c=='-')t=-1;
    do{s=s*10+c-'0';}while(isdigit(c=getchar()));
    return s*t;
}
inline ll readl(){
    char c;ll s=0;int t=1;
    while(!isdigit(c=getchar()))if(c=='-')t=-1;
    do{s=s*10+c-'0';}while(isdigit(c=getchar()));
    return s*t;
}
/*------------------------------------------------------------*/
#define mod 1000000007
#define maxn 200010
#define inf 0x7fffffff
/*------------------------------------------------------------*/
ll a[maxn],b[maxn];
int main(){
	ll n;
	inl(n);
	ll ans=0;
	ll k=0;
	FOR(i,1,n){
		inl2(a[i],b[i]);
		k=max(k+b[i]-(a[i]-a[i-1]),b[i]);
		ans=max(ans,k);
	}
	outl(ans);
}
### Prompt

Your task is to create 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
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define il inline
#define outy printf("Yes\n")
#define outn printf("No\n")
#define outf printf("First\n")
#define outs printf("Second\n")
#define FOR(a,b,c) for(int a=b;a<=c;a++)
#define FORD(a,b,c) for(int a=b;a>=c;a--)
#define FORL(a,b) for(int a=head[b];a;a=nxt[a])
#define in(a)  a=read()
#define in2(a,b)  a=read(),b=read()
#define in3(a,b,c)  a=read(),b=read(),c=read()
#define in4(a,b,c,d)  a=read(),b=read(),c=read(),d=read()
#define inl(a)  a=readl()
#define inl2(a,b)  a=readl(),b=readl()
#define inl3(a,b,c)  a=readl(),b=readl(),c=readl()
#define inl4(a,b,c,d)  a=readl(),b=readl(),c=readl(),d=readl()
#define out(a) printf("%d\n",a)
#define out2(a,b) printf("%d %d\n",a,b)
#define out3(a,b,c) printf("%d %d %d\n",a,b,c)
#define out4(a,b,c,d) printf("%d %d %d %d\n",a,b,c,d)
#define outl(a) printf("%lld\n",a)
#define outl2(a,b) printf("%lld %lld\n",a,b)
#define outl3(a,b,c) printf("%lld %lld %lld\n",a,b,c)
#define outl4(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)
#define clr(x,y) memset(x,y,sizeof(x))
inline int read(){
    char c;int s=0,t=1;
    while(!isdigit(c=getchar()))if(c=='-')t=-1;
    do{s=s*10+c-'0';}while(isdigit(c=getchar()));
    return s*t;
}
inline ll readl(){
    char c;ll s=0;int t=1;
    while(!isdigit(c=getchar()))if(c=='-')t=-1;
    do{s=s*10+c-'0';}while(isdigit(c=getchar()));
    return s*t;
}
/*------------------------------------------------------------*/
#define mod 1000000007
#define maxn 200010
#define inf 0x7fffffff
/*------------------------------------------------------------*/
ll a[maxn],b[maxn];
int main(){
	ll n;
	inl(n);
	ll ans=0;
	ll k=0;
	FOR(i,1,n){
		inl2(a[i],b[i]);
		k=max(k+b[i]-(a[i]-a[i-1]),b[i]);
		ans=max(ans,k);
	}
	outl(ans);
}
```