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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 
public class Solution{
static FastScanner sc = new FastScanner();
public static void solve(){
    int n=sc.nextInt();
    long[] a=sc.readLongArray(n);
    char[] c=sc.next().toCharArray();
    for (int i=0;i<n;i++) {
      if (a[i]<=0 && c[i]=='B') {
        System.out.println("NO");
        return;
      }
      if (a[i]>n && c[i]=='R') {
        System.out.println("NO");
        return;
      }
    }

    List<Long> lst1=new ArrayList<>();
    List<Long> lst2=new ArrayList<>();
    for (int i=0;i<n;i++) {
      if (c[i]=='R') {
        lst1.add(a[i]);
      }else
      lst2.add(a[i]);
    }
    Collections.sort(lst1);
    Collections.sort(lst2);
    int i=0,j=0,ct=1;
    int s1=lst1.size();
    int s2=lst2.size();
    while(i<s1 && j<s2){
      if(lst2.get(j)<ct){
        System.out.println("NO");
        return;
      }
      if (lst1.get(i)>ct && lst2.get(j)<ct) {
        System.out.println("NO");
        return;
      }else if(lst1.get(i)==ct && lst2.get(j)==ct){
          j++;
      }else if(lst2.get(j)==ct ){
        j++;
      }else if(lst1.get(i)==ct ){
        i++;
      }else{
        j++;
      }
      ct++;
    }
    if (j<s2) {
      while (j<s2) {
        if (lst2.get(j)>=ct) {
          ct++;
        }
        else break;
        j++;
      }
    }
    if (j!=s2) {
      System.out.println("NO");
      return;
    }
    if (i<s1) {
      // System.out.println("IN");
      while (i<s1) {
        if (lst1.get(i)<=ct) {
          ct++;
        }else
        break;
        i++;
      }
    }
    if (i!=s1) {
      System.out.println("NO");
      return;
    }
    System.out.println("YES");
    }


public static void main(String[] args) {
  int t = sc.nextInt();
  outer: for (int tt = 0; tt < t; tt++) {
    solve();
  }

}
static boolean isPrime(int n)
    {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
  
        if (n % 2 == 0 || n % 3 == 0)
            return false;
  
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
  
        return true;
    }
public static int[] LPS(String s){
  int[] lps=new int[s.length()];
  int i=0,j=1;
  while (j<s.length()) {
    if(s.charAt(i)==s.charAt(j)){
      lps[j]=i+1;
      i++;
      j++;
      continue;
    }else{
      if (i==0) {
        j++;
        continue;
      }
      i=lps[i-1];
      while(s.charAt(i)!=s.charAt(j) && i!=0) {
        i=lps[i-1];
      }
      if(s.charAt(i)==s.charAt(j)){
      lps[j]=i+1;
      i++;
      }
      j++;
    }
  }
  return lps;
}
static long getPairsCount(int n, double sum,int[] arr)
    {
        HashMap<Double, Integer> hm = new HashMap<>();
        for (int i = 0; i < n; i++) {
            if (!hm.containsKey((double)arr[i]))
                hm.put((double)arr[i], 0);
            hm.put((double)arr[i], hm.get((double)arr[i]) + 1);
        }
        long twice_count = 0;
        for (int i = 0; i < n; i++) {
            if (hm.get(sum - arr[i]) != null)
                twice_count += hm.get(sum - arr[i]);
            if (sum - (double)arr[i] == (double)arr[i])
                twice_count--;
        }
        return twice_count / 2l;
    }
static boolean[] sieveOfEratosthenes(int n)
    {
        boolean prime[] = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;
 
        for (int p = 2; p * p <= n; p++)
        {
            if (prime[p] == true)
            {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;
 }
static long power(long x, long y, long p)
  {
    long res = 1l; 
    x = x % p;
    if (x == 0)
      return 0;
    while (y > 0)
    {

      if ((y & 1) != 0)
        res = (res * x) % p;
      y>>=1;
      x = (x * x) % p;
    }
    return res;
  }

public static int log2(int N)
    {
        int result = (int)(Math.log(N) / Math.log(2));
  
        return result;
    }

////////////////////////////////////////////////////////////////////////////////////
////////////////////DO NOT READ AFTER THIS LINE //////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////
static long modFact(int n, 
                       int p) 
    { 
        if (n >= p) 
            return 0; 
      
        long result = 1l; 
        for (int i = 3; i <= n; i++) 
            result = (result * i) % p; 
      
        return result; 
    } 
      
static boolean isPalindrom(char[] arr, int i, int j) {
  boolean ok = true;
  while (i <= j) {
    if (arr[i] != arr[j]) {
      ok = false;
      break;
    }
    i++;
    j--;
  }
  return ok;
}

static int max(int a, int b) {
  return Math.max(a, b);
}

static int min(int a, int b) {
  return Math.min(a, b);
}

static long max(long a, long b) {
  return Math.max(a, b);
}

static long min(long a, long b) {
  return Math.min(a, b);
}

static int abs(int a) {
  return Math.abs(a);
}

static long abs(long a) {
  return Math.abs(a);
}

static void swap(long arr[], int i, int j) {
  long temp = arr[i];
  arr[i] = arr[j];
  arr[j] = temp;
}

static void swap(int arr[], int i, int j) {
  int temp = arr[i];
  arr[i] = arr[j];
  arr[j] = temp;
}

static int maxArr(int arr[]) {
  int maxi = Integer.MIN_VALUE;
  for (int x : arr)
    maxi = max(maxi, x);
  return maxi;
}

static int minArr(int arr[]) {
  int mini = Integer.MAX_VALUE;
  for (int x : arr)
    mini = min(mini, x);
  return mini;
}

static long maxArr(long arr[]) {
  long maxi = Long.MIN_VALUE;
  for (long x : arr)
    maxi = max(maxi, x);
  return maxi;
}

static long minArr(long arr[]) {
  long mini = Long.MAX_VALUE;
  for (long x : arr)
    mini = min(mini, x);
  return mini;
}
static int lcm(int a,int b){
  return (int)(((long)a*b)/(long)gcd(a,b));
}
static int gcd(int a, int b) {
  if (b == 0)
    return a;
  return gcd(b, a % b);
}

static long gcd(long a, long b) {
  if (b == 0)
    return a;
  return gcd(b, a % b);
}

static void ruffleSort(int[] a) {
  int n = a.length;
  Random r = new Random();
  for (int i = 0; i < a.length; i++) {
    int oi = r.nextInt(n);
    int temp = a[i];
    a[i] = a[oi];
    a[oi] = temp;
  }
  Arrays.sort(a);
}

public static int binarySearch(int a[], int target) {
  int left = 0;
  int right = a.length - 1;
  int mid = (left + right) / 2;
  int i = 0;
  while (left <= right) {
    if (a[mid] <= target) {
      i = mid + 1;
      left = mid + 1;
    } else {
      right = mid - 1;
    }
    mid = (left + right) / 2;
  }
  return i-1;
}


static class FastScanner {
  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  StringTokenizer st = new StringTokenizer("");

  String next() {
    while (!st.hasMoreTokens())
      try {
        st = new StringTokenizer(br.readLine());
      } catch (IOException e) {
        e.printStackTrace();
      }
    return st.nextToken();
  }

  int nextInt() {
    return Integer.parseInt(next());
  }

  int[] readArray(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = nextInt();
    return a;
  }

  long[] readLongArray(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nextLong();
    return a;
  }

  int[][] read2dArray(int n, int m) {
    int arr[][] = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        arr[i][j] = nextInt();
      }
    }
    return arr;
  }

  ArrayList<Integer> readArrayList(int n) {
    ArrayList<Integer> arr = new ArrayList<Integer>();
    for (int i = 0; i < n; i++) {
      int a = nextInt();
      arr.add(a);
    }
    return arr;
  }

  long nextLong() {
    return Long.parseLong(next());
  }
}
 static class Pair {
  int l, r;

  Pair(int fr, int sc) {
    this.l = fr;
    this.r = sc;
  }
}
}