codeforces 669B B. Little Artem and Grasshopper(水题)

题目链接：

B. Little Artem and Grasshopper

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Artem found a grasshopper. He brought it to his house and constructed a jumping area for him.

The area looks like a strip of cells 1 × n. Each cell contains the direction for the next jump and the length of that jump. Grasshopper starts in the first cell and follows the instructions written on the cells. Grasshopper stops immediately if it jumps out of the strip. Now Artem wants to find out if this will ever happen.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — length of the strip.

Next line contains a string of length n which consists of characters "<" and ">" only, that provide the direction of the jump from the corresponding cell. Next line contains n integers di (1 ≤ di ≤ 109) — the length of the jump from the i-th cell.

Output

Print "INFINITE" (without quotes) if grasshopper will continue his jumps forever. Otherwise print "FINITE" (without quotes).

Examples

input

2
><
1 2

output

FINITE

input

3
>><
2 1 1

output

INFINITE

Note
In the first sample grasshopper starts from the first cell and jumps to the right on the next cell. When he is in the second cell he needs to jump two cells left so he will jump out of the strip.

Second sample grasshopper path is 1 - 3 - 2 - 3 - 2 - 3 and so on. The path is infinite

题意:

给每个格子的跳动方向和跳动的距离,问从第一个格子开始跳,是永远在这些格子里还是会跳出去;

思路:

模拟跳动,跳出去了就输出跳出去了,没跳出去当跳到原来跳过的格子上就会永远跳不出去;

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll inf=1e15;
const int N=1e5+6;
int n,a
,vis
;
char s
;
int main()
{
memset(vis,0,sizeof(vis));
scanf("%d",&n);
scanf("%s",s+1);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int pos=1;
vis[1]=1;
while(1)
{
if(s[pos]=='>')
{
if(pos+a[pos]>n)
{
printf("FINITE\n");
return 0;
}
else
{
if(vis[pos+a[pos]])
{
printf("INFINITE\n");
return 0;
}
else
{
pos+=a[pos];
vis[pos]=1;
}
}
}
else
{

if(pos-a[pos]<1)
{
printf("FINITE\n");
return 0;
}
else
{
if(vis[pos-a[pos]])
{
printf("INFINITE\n");
return 0;
}
else
{
pos-=a[pos];
vis[pos]=1;
}
}

}
}

return 0;
}