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.
 //package eround101;

import java.util.*;
import java.io.*;
import java.lang.*;
import java.util.StringTokenizer;

public class Solution {

    static HritikScanner sc = new HritikScanner();
    static PrintWriter pw = new PrintWriter(System.out, true);
    final static int MOD = 1000000007;
    static StringBuilder sb = new StringBuilder();

    public static void main(String[] args) {

        int t = ni();
        while (t-- > 0) {
            solve();
        }
    }

    static void solve() {
        int n = ni();
        int[] arr = nextIntArray(n);
        char[] col = sc.next().toCharArray();
        int[] cB = new int[n+1];
        int[] cR = new int[n+1];
        for(int i = 0; i < n; i++)
        {
//            pl((col[i] == 'R' && arr[i] > n));
//            pl((col[i] == 'B' && arr[i] < 1));
            if((col[i] == 'R' && arr[i] > n)||(col[i] == 'B' && arr[i] < 1))
            {
                pl("NO");
                return;
            }
            if(col[i] == 'B')
            {
                if(arr[i] > n)
                    continue;
                cB[arr[i]]++;
                if(cB[arr[i]] > arr[i])
                {
                    pl("NO");
                }
            }
            else
            {
                if(arr[i] < 1)
                    continue;
                cR[arr[i]]++;
                if(cR[arr[i]] > (n-(arr[i]-1)))
                {
                    pl("NO");
                }
            }
//            pa(cB);
//            pa(cR);
        }
        int[] psum = new int[n+1];
        for(int i = 1; i<= n; i++)
        {
            psum[i] = psum[i-1]+cB[i];
            if(psum[i] > i)
            {
                pl("NO");
                return;
            }
        }
        int[] psum1 = new int[n+1];
        psum1[n] = cR[n];
        for(int i = n-1; i>= 0; i--)
        {
            psum1[i] = psum[i+1]+cR[i];
            if(psum1[i] > (n-(i-1)))
            {
                pl("NO");
                return;
            }
        }
        
        pl("YES");
        
    }
        /////////////////////////////////////////////////////////////////////////////////
    static class FenwickTree { // Binary Index Tree
        int[] tree;
        static int size;

        public FenwickTree(int size) {
            this.size = size;
            tree = new int[size + 5];
        }

        public void add(int i, int val) {
            while (i <= size) {
                tree[i] += val;
                i += i & -i;  // adding the decimal value of the last set bit.
            }
        }

        public int sum(int i) {
            int res = 0;
            while (i >= 1) {
                res += tree[i];
                i -= i & -i; // deleting the last set bit
            }
            return res;
        }

        public int sum(int l, int r) {
            return sum(r) - sum(l - 1);
        }
    }

    /////////////////////////////////////////////////////////////////////////////////
    static int[] nextIntArray(int n) {
        int[] arr = new int[n];
        int i = 0;
        while (i < n) {
            arr[i++] = ni();
        }
        return arr;
    }

    static long[] nextLongArray(int n) {
        long[] arr = new long[n];
        int i = 0;
        while (i < n) {
            arr[i++] = nl();
        }
        return arr;
    }

    static int[] nextIntArray1(int n) {
        int[] arr = new int[n + 1];
        int i = 1;
        while (i <= n) {
            arr[i++] = ni();
        }
        return arr;
    }

    /////////////////////////////////////////////////////////////////////////////////
    static int ni() {
        return sc.nextInt();
    }

    static long nl() {
        return sc.nextLong();
    }

    static double nd() {
        return sc.nextDouble();
    }

    /////////////////////////////////////////////////////////////////////////////////
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o + " ");
    }

    static void pl(Object o) {
        pw.println(o);
    }

    static void psb(StringBuilder sb) {
        pw.print(sb);
    }

    static void pa(Object arr[]) {
        for (Object o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(int arr[]) {
        for (int o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(long arr[]) {
        for (long o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(double arr[]) {
        for (double o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(char arr[]) {
        for (char o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(List list) {
        for (Object o : list) {
            p(o);
        }
        pl();
    }

    static void pa(Object[][] arr) {

        for (int i = 0; i < arr.length; ++i) {
            for (Object o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(int[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (int o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(long[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (long o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(char[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (char o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(double[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (double o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

/////////////////////////////////////////////////////////////////////////////////
    static void print(Object s) {
        System.out.println(s);
    }

    /////////////////////////////////////////////////////////////////////////////////
    //-----------HritikScanner class for faster input----------//
    static class HritikScanner {

        BufferedReader br;
        StringTokenizer st;

        public HritikScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    //////////////////////////////////////////////////////////////////
    public static class Pair implements Comparable<Pair> {

        int num, row, col;

        Pair(int num, int row, int col) {
            this.num = num;
            this.row = row;
            this.col = col;
        }

        public int get1() {
            return num;
        }

        public int get2() {
            return row;
        }

        public int compareTo(Pair A) {
            return this.num - A.num;
        }

        public String toString() {
            return num + " " + row + " " + col;
        }
    }

    //////////////////////////////////////////////////////////////////
    // Function to return gcd of a and b time complexity O(log(a+b))
    static long gcd(long a, long b) {
        if (a == 0) {
            return b;
        }
        return gcd(b % a, a);
    }

    // method to return LCM of two numbers 
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    //////////////////////////////////////////////////////////////////
    static boolean isPrime(long n) {
        // Corner cases 
        if (n <= 1) {
            return false;
        }
        if (n <= 3) {
            return true;
        }

        // This is checked so that we can skip  
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) {
            return false;
        }

        for (long i = 5; i * i <= n; i = i + 6) {
            if (n % i == 0 || n % (i + 2) == 0) {
                return false;
            }
        }

        return true;
    }
    //////////////////////////////////////////////////////////////////

    static boolean isPowerOfTwo(long n) {
        if (n == 0) {
            return false;
        }

        return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
    }

    public static long getFact(int n) {
        long ans = 1;
        while (n > 0) {
            ans *= n;
            ans %= MOD;
            n--;
        }
        return ans;
    }

    public static long pow(long n, int pow) {
        if (pow == 0) {
            return 1;
        }
        long temp = pow(n, pow / 2) % MOD;
        temp *= temp;
        temp %= MOD;
        if (pow % 2 == 1) {
            temp *= n;
        }
        temp %= MOD;
        return temp;

    }

    public static long nCr(int n, int r) {
        long ans = 1;
        int temp = n - r;
        while (n > temp) {
            ans *= n;
            ans %= MOD;
            n--;
        }
        ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
        ans %= MOD;
        return ans;
    }

    //////////////////////////////////////////////////////////////////
    // method returns Nth power of A 
    static double nthRoot(int A, int N) {

        // intially guessing a random number between 
        // 0 and 9 
        double xPre = Math.random() % 10;

        // smaller eps, denotes more accuracy 
        double eps = 0.001;

        // initializing difference between two 
        // roots by INT_MAX 
        double delX = 2147483647;

        // xK denotes current value of x 
        double xK = 0.0;

        // loop untill we reach desired accuracy 
        while (delX > eps) {
            // calculating current value from previous 
            // value by newton's method 
            xK = ((N - 1.0) * xPre
                    + (double) A / Math.pow(xPre, N - 1)) / (double) N;
            delX = Math.abs(xK - xPre);
            xPre = xK;
        }

        return xK;
    }

}
