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You have an array a consisting of n integers. Each integer from 1 to n appears exactly once in this array.

For some indices i (1 ≤ i ≤ n - 1) it is possible to swap i-th element with (i + 1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap i-th element with (i + 1)-th (if the position is not forbidden).

Can you make this array sorted in ascending order performing some sequence of swapping operations?

Input

The first line contains one integer n (2 ≤ n ≤ 200000) — the number of elements in the array.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 200000) — the elements of the array. Each integer from 1 to n appears exactly once.

The third line contains a string of n - 1 characters, each character is either 0 or 1. If i-th character is 1, then you can swap i-th element with (i + 1)-th any number of times, otherwise it is forbidden to swap i-th element with (i + 1)-th.

Output

If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.

Examples

Input

6
1 2 5 3 4 6
01110

Output

YES

Input

6
1 2 5 3 4 6
01010

Output

NO

Note

In the first example you may swap a₃ and a₄, and then swap a₄ and a₅.

题意：

给定1~n的序列（所有数刚好出现一次），然后再给01串，01串中如果是1，则可以交换当前位置和下一个位置的数字。问是否能还原为递增数组。

思路：

如果当前数的01串对应是0，且前一个位置01串也是0，且当前数字与下标不对应，肯定输出NO，因为没法交换

然后对可以交换的部分记录下最大最小值，然后记录下起始位置和终点位置，比较一下就行了。

比如说第一组数据

6
1 2 5 3 4 6
01110

01串第一个是0，对应上去是1，下标也是1，则可以。

然后从2~4都是可以交换的范围（即连续的1，然后后面第一个0）

保留下最大值是4，最小值是2，起始位置2，末尾位置4。一一对应了，就输出YES。

代码：

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
int read(){
    char c; bool op = 0;
    while((c = getchar()) < '0' || c > '9')
        if(c == '-') op = 1;
    int res = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        res = res * 10 + c - '0';
    return op ? -res : res;
}
const int maxn = 1e6+7;
int a[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1 ; i <= n ; i++)
        scanf("%d",&a[i]);
    int Max = 0,Min = n+1;
    int flag = 0,ardMin;
    getchar();
    for(int i = 1 ; i <= n - 1 ; i++){
        char c;
        scanf("%c",&c);
        if(flag == 0 && c == '1'){
            ardMin = i;
        }
        if(c == '0' && flag == 1){
            Max = max(Max,a[i]);
            Min = min(Min,a[i]);
            if(ardMin!= Min || Max != i){
                return puts("NO"),0;
            }
            Max = 0;Min = n+1;
            flag = 0;
        }
        else if(c == '1'){
            Max = max(Max,a[i]);
            Min = min(Min,a[i]);
            flag = 1;
        }
        else if(c == '0'&& a[i] != i){
            return puts("NO"),0;
        }
    }
    return puts("YES"),0;
}
/*
6
1 3 2 4 6 5
01000
*/