You are given an array of integers a of length n. The elements of the array can be either different or the same. 

Each element of the array is colored either blue or red. There are no unpainted elements in the array. One of the two operations described below can be applied to an array in a single step:

  * either you can select any blue element and decrease its value by 1; 
  * or you can select any red element and increase its value by 1. 



Situations in which there are no elements of some color at all are also possible. For example, if the whole array is colored blue or red, one of the operations becomes unavailable.

Determine whether it is possible to make 0 or more steps such that the resulting array is a permutation of numbers from 1 to n?

In other words, check whether there exists a sequence of steps (possibly empty) such that after applying it, the array a contains in some order all numbers from 1 to n (inclusive), each exactly once.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of input data sets in the test.

The description of each set of input data consists of three lines. The first line contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the original array a. The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the array elements themselves.

The third line has length n and consists exclusively of the letters 'B' and/or 'R': ith character is 'B' if a_i is colored blue, and is 'R' if colored red.

It is guaranteed that the sum of n over all input sets does not exceed 2 ⋅ 10^5.

Output

Print t lines, each of which contains the answer to the corresponding test case of the input. Print YES as an answer if the corresponding array can be transformed into a permutation, and NO otherwise.

You can print the answer in any case (for example, the strings yEs, yes, Yes, and YES will be recognized as a positive answer).

Example

Input


8
4
1 2 5 2
BRBR
2
1 1
BB
5
3 1 4 2 5
RBRRB
5
3 1 3 1 3
RBRRB
5
5 1 5 1 5
RBRRB
4
2 2 2 2
BRBR
2
1 -2
BR
4
-2 -1 4 0
RRRR


Output


YES
NO
YES
YES
NO
YES
YES
YES

Note

In the first test case of the example, the following sequence of moves can be performed:

  * choose i=3, element a_3=5 is blue, so we decrease it, we get a=[1,2,4,2]; 
  * choose i=2, element a_2=2 is red, so we increase it, we get a=[1,3,4,2]; 
  * choose i=3, element a_3=4 is blue, so we decrease it, we get a=[1,3,3,2]; 
  * choose i=2, element a_2=2 is red, so we increase it, we get a=[1,4,3,2]. 



We got that a is a permutation. Hence the answer is YES.
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 2e5 + 5;
const long long SQRTN = 1003;
const long long LOGN = 22;
const double PI = acos(-1);
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const long long FMOD = 998244353;
const double eps = 1e-9;
void __print(long long x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
long long add(long long a, long long b, long long c = MOD) {
  long long res = a + b;
  return (res >= c ? res % c : res);
}
long long sub(long long a, long long b, long long c = MOD) {
  long long res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
long long mul(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
long long muln(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return ((res % c) + c) % c;
}
long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T binpow(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T binpow2(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T &x, T &y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : sub(x, z, n));
}
const long long FACSZ = 1e4;
long long fact[FACSZ], ifact[FACSZ];
void precom(long long c = MOD) {
  fact[0] = 1;
  for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
  ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
  for (long long i = FACSZ - 1 - 1; i >= 0; i--) {
    ifact[i] = mul(i + 1, ifact[i + 1], c);
  }
}
vector<long long> primes;
void prime_precom() {
  primes.push_back(2);
  for (long long x = 3; primes.size() <= MAXN; x += 2) {
    bool isPrime = true;
    for (auto p : primes) {
      if (x % p == 0) {
        isPrime = false;
        break;
      }
      if (p * p > x) {
        break;
      }
    }
    if (isPrime) {
      primes.push_back(x);
    }
  }
}
long long ncr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long ncr_modp(long long n, long long k, long long c = MOD) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res = mul(res, n - i, c);
    res = mul(res, binpow2(i + 1, c - 2, c), c);
  }
  return res;
}
vector<long long> factors;
void factorize(long long a) {
  factors.clear();
  for (long long i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      factors.push_back(i);
      factors.push_back(a / i);
    }
  }
  sort(factors.begin(), factors.end());
}
long long ncr_precom(long long n, long long r, long long c = MOD) {
  return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
}
long long ceil(long long a, long long b) { return (a + b - 1) / b; }
bool is_prime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
bool diophantine_checker(long long a, long long b, long long n) {
  for (long long i = 0; i * a <= n; i++) {
    if ((n - (i * a)) % b == 0) {
      return true;
    }
  }
  return false;
}
long long count_divisors(long long n) {
  long long c;
  long long ans = 1;
  for (long long i = 2; i * i <= n; i++) {
    c = 0;
    while (n % i == 0) {
      c++;
      n /= i;
    }
    ans *= (c + 1);
  }
  if (n > 2) {
    return ans * 2;
  }
  return ans;
}
string to_binary(long long n) {
  string r;
  while (n != 0) {
    r = (n % 2 == 0 ? "0" : "1") + r;
    n /= 2;
  }
  return r;
}
bool ispower2(long long x) { return x && (!(x & (x - 1))); }
vector<long long> spf;
void sieve() {
  spf.resize(MAXN);
  spf[1] = 1;
  for (long long i = 2; i < MAXN; i++) {
    spf[i] = i;
  }
  for (long long i = 4; i < MAXN; i += 2) {
    spf[i] = 2;
  }
  for (long long i = 3; i * i <= MAXN; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < MAXN; j += i) {
        if (spf[j] == j) {
          spf[j] = i;
        }
      }
    }
  }
}
void pfactor(long long x, vector<long long> &ret) {
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
}
void solvethetestcase() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  string s;
  cin >> s;
  vector<long long> cnt(2, 0ll);
  for (long long i = 0; i < n; i++) {
    if (s[i] == 'R') {
      if (v[i] == n) {
        cnt[1]++;
      }
    } else {
      if (v[i] == 1) {
        cnt[0]++;
      }
    }
  }
  if (cnt[1] >= 2 || cnt[0] > 1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
  }
}
signed main() {
  cout << fixed << setprecision(12);
  ;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long testcase = 1; testcase < t + 1; testcase++) {
    solvethetestcase();
  }
}
