cf - 920 c 求能否实现交换

C. Swap Adjacent Elements

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题目分析 ： 一串数字，还有一串字符串，为 1 的时候当前的数字可以和它下一位去交换，然后这么想，对于一个不在当前位置的数字，如果它大于当前的位置，那么它如果想要被交换到相应的位置，那么字符串的当前位置和要交换到的位置的前一个字符就应该都是 1 ，那么前缀和去处理字符串就可以了，还有就是只需要考虑数大于当前位置的情况，大的满足，小的一定成立。

代码示例 ：

const int eps = 2e5+5;
const double pi = acos(-1.0);
const int inf = 1<<29;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long

int pre[eps];
char s[eps];
int arr[eps];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n;
    
    cin >> n;
    for(int i = 1; i <= n; i++){
        scanf("%d", &pre[i]);
    }
    scanf("%s", s+1);
    for(int i = 1; i < n; i++){
        if (s[i] == '1') arr[i] = 1;
        else arr[i] = 0;
        arr[i] += arr[i-1];
    }
   //for(int i = 1; i < n; i++){
    //printf("%d ", arr[i]);
   //} 
   //printf("\n");
    int sign = 0;
    for(int i = 1; i < n; i++){
        if (pre[i] > i) {
            int f = arr[pre[i]-1] - arr[i-1];
            if (f != pre[i]-i) {sign = 1; break;}
        }
        if (sign) break;
    }
    if (!sign) printf("YES\n");
    else printf("NO\n");
    return 0;
}