Educational Codeforces Round 37 (Rated for Div. 2)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

Copy

61 2 5 3 4 601110

output

Copy

YES

input

Copy

61 2 5 3 4 601010

output

Copy

NO

Note

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

题意：

给你一个1~n的全排列数组a，

以及一个数组can[i] ，如果can[i] =1 则表示 a[i] 可以和a[i+1] 交换，

每个位置都可以交换任意次（包括0次）

问你是否可以通过一些交换operation使整个数组是有序的。

思路：

求下can数组的前缀和 sum，

数组a中，a[i] 一定要移动到 i 才可以让整个数组有序。

那么我们只需要判断 当 \(a[i]>i\) 时，$ [i,a[i]-1 ] $ 这个区间中can是否全为1。

显然有前缀和判断即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int can[maxn];
int a[maxn];
int n;
int sum[maxn];
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    du1(n);
    repd(i, 1, n) {
        du1(a[i]);
    }
    repd(i, 1, n - 1) {
        scanf("%1d", &can[i]);
        sum[i] = sum[i - 1] + can[i];
    }
    int isok = 1;
    repd(i, 1, n - 1) {
        if (a[i] > i) {
            if (sum[a[i] - 1] - sum[i - 1] == a[i]-i) {

            } else {
                isok = 0;
            }
        }
    }
    if (isok) {
        puts("YES");
    } else {
        puts("NO");
    }

    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}