Example

Input

3
y 7
y 6
n 5


Output

1
// This amazing code is by Eric Sunli Chen.
#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <vector>
using namespace std;
template<typename T> void get_int(T &x)
{
	char t=getchar();
	bool neg=false;
	x=0;
	for(; (t>'9'||t<'0')&&t!='-'; t=getchar());
	if(t=='-')neg=true,t=getchar();
	for(; t<='9'&&t>='0'; t=getchar())x=x*10+t-'0';
	if(neg)x=-x;
}
template<typename T> void print_int(T x)
{
	if(x<0)putchar('-'),x=-x;
	short a[20]= {},sz=0;
	while(x>0)a[sz++]=x%10,x/=10;
	if(sz==0)putchar('0');
	for(int i=sz-1; i>=0; i--)putchar('0'+a[i]);
}
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define get1(a) get_int(a)
#define get2(a,b) get1(a),get1(b)
#define get3(a,b,c) get1(a),get2(b,c)
#define printendl(a) print_int(a),puts("")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const LL Linf=1ll<<61;
const double pi=acos(-1.0);

int n,len[1111],ok[1111],mn[1111][1111];
map<int,int> tt[1111][1111];
char tmp[5];
int dfs(int l,int r,int up)
{
	if(l>r)return 0;
	if(tt[l][r].find(up)!=tt[l][r].end())return tt[l][r][up];
	int mx=0,tot=0,cnt=0;
	for(int i=l;i<=r;i++)
	{
		if(!ok[i])
		{
			if(mx==0||len[mx]<len[i])mx=i;
			if(cnt)
			{
				tot++;
				cnt=0;
			}
		}
		else if(ok[i]&&len[i]<=up)cnt++;
	}
	if(cnt)tot++;
	if(tot==0)return tt[l][r][up]=0;
	if(mx==0)return tt[l][r][up]=tot;
//	printf("%d %d %d tot= %d mx= %d\n",l,r,up,tot);
	return tt[l][r][up]=min(dfs(l,mx-1,len[mx])+dfs(mx+1,r,len[mx])+1,min(dfs(l,mx-1,up)+dfs(mx+1,r,up),tot));
}
int main()
{
	get1(n);
	for(int i=1;i<=n;i++)
	{
		scanf("%s%d",tmp,len+i);
		if(tmp[0]=='y')ok[i]=1;
	}
	printendl(dfs(1,n,inf));
	return 0;
}