【codeforces 750B】New Year and North Pole

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.

Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: “North”, “South”, “West”, “East”.

Limak isn’t sure whether the description is valid. You must help him to check the following conditions:

If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.

If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.

The journey must end on the North Pole.

Check if the above conditions are satisfied and print “YES” or “NO” on a single line.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 50).

The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, ) — the length and the direction of the i-th part of the journey, according to the description Limak got.

Output

Print “YES” if the description satisfies the three conditions, otherwise print “NO”, both without the quotes.

Examples

input

5

7500 South

10000 East

3500 North

4444 West

4000 North

output

YES

input

2

15000 South

4000 East

output

NO

input

5

20000 South

1000 North

1000000 West

9000 North

10000 North

output

YES

input

3

20000 South

10 East

20000 North

output

NO

input

2

1000 North

1000 South

output

NO

input

4

50 South

50 North

15000 South

15000 North

output

YES

Note

Drawings below show how Limak’s journey would look like in first two samples. In the second sample the answer is “NO” because he doesn’t end on the North Pole.

【题目链接】:http://codeforces.com/contest/750/problem/B

【题解】



细节题

当前的位置只要记录横纵坐标就可以了;

一开始纵坐标位置为20000

对于左右的处理

其他情况下都不用管

如果有一个向上或向下

但是它的值大于20000,则也直接输出no

如果位置x+t>20000或x-t<0也直接输出no

如果当前位置是20000,则如果操作不是往下也直接输出NO

如果当前位置是0,如果操作不是往上则也直接输出NO

最后判断当前的位置是不是20000



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n;
LL x;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    x = 20000;
    rei(n);
    rep1(i,1,n)
    {
        LL ti;
        string s;
        cin >> ti >> s;
        if (x==20000)
        {
            if (s[0]!='S')
            {
                puts("NO");
                return 0;
            }
        }
        if (x==0)
        {
            if (s[0]!='N')
            {
                puts("NO");
                return 0;
            }
        }
        if (s[0]=='S'||s[0]=='N')
        {
            if (ti>20000)
            {
                puts("NO");
                return 0;
            }
        }
        if (s[0]=='S')
        {
            if (x-ti<0)
            {
                puts("NO");
                return 0;
            }
            else
                x-=ti;
        }
        if (s[0]=='N')
        {
            if (x+ti>20000)
            {
                puts("NO");
                return 0;
            }
            else
                x+=ti;
        }
    }
    if (x==20000)
        puts("YES");
    else
        puts("NO");
    return 0;
}