Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) B

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 d_i (1 ≤ d_i ≤ 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：有n个区域每个区域给一个固定跳跃方向‘>’代表向右和 ‘<’代表向左接下来的一行代表这个从这个区域能跳跃多远

若能一直跳跃输出INFINITE 否则输出FINITE

题解：存储每个位置的跳跃后的状态暴力模拟过程

标记若在某一个位置重复出现则说明能够一直跳跃输出INFINITE 若跳出1~n的范围则输出FINITE

 1 #include<iostream>
 2  #include<cstring>
 3  #include<cstdio>
 4  #include<queue>
 5  #include<stack>
 6  #include<map>
 7  #include<set>
 8  #include<algorithm>
 9  #define ll __int64
10  #define pi acos(-1.0)
11  #define mod 1
12  #define maxn 10000
13  using namespace std;
14  int n;
15  map<int,char> mp;
16  map<int,int> to;
17  map<int,int> mpp;
18  int exm;
19  int main()
20  {
21      mp.clear();
22      mpp.clear();
23      to.clear();
24      scanf("%d",&n);
25      getchar();
26      for(int i=1;i<=n;i++)
27         scanf("%c",&mp[i]);
28      for(int i=1;i<=n;i++)
29      {
30          scanf("%d",&exm);
31          if(mp[i]=='>')
32            to[i]=i+exm;
33          else
34            to[i]=i-exm;
35     }
36     exm=1;
37     int flag=0;
38     while(1)
39     {
40         if(exm>n||exm<1)
41         {
42             flag=0;
43             break;
44         }
45         if(mpp[exm]==1)
46         {
47             flag=1;
48             break;
49         }
50         mpp[exm]=1;
51         exm=to[exm];    
52     }
53     if(flag)
54    cout<<"INFINITE"<<endl; 
55    else
56     cout<<"FINITE"<<endl;    
57      return 0;
58  }