B. Phoenix and Beauty

// Problem: B. Trouble Sort
// Contest: Codeforces - Codeforces Round #648 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1365/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-01 15:43:40
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
void solve() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n ;i ++) {
		cin >> a[i];
	}
	int ans = 0;
	for (int i = 0; i < n; i ++){
		int t;
		cin >> t;
		ans += t;
	}
	if (ans && ans < n || is_sorted(all(a))) puts("Yes");
	else puts("No");
}
int main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

as long as 0 and 1 If the number of is different, we can put them in positive order . Because for every number that is not in position , We can put him in the right place b The opposite number of , Then exchange with the number in the correct position Regardless of the specific number of positions