#include <bits/stdc++.h>
using namespace std;
int main() {
  int tests;
  scanf("%d", &tests);
  for (int test = 0; test < (int)tests; test++) {
    int n;
    scanf("%d", &n);
    vector<int> a(n);
    for (int i = 0; i < (int)n; i++) {
      scanf("%d", &a[i]);
    }
    string s;
    cin >> s;
    multiset<int> red, blue;
    for (int i = 0; i < (int)n; i++) {
      if (s[i] == 'B') {
        blue.insert(a[i]);
      } else {
        red.insert(a[i]);
      }
    }
    bool fail = false;
    for (int i = (int)1; i <= (int)n; i++) {
      while (!blue.empty() && *blue.begin() < i) {
        blue.erase(blue.begin());
      }
      if (!blue.empty()) {
        blue.erase(blue.begin());
      } else {
        auto iter = red.upper_bound(i);
        if (iter == red.begin()) {
          fail = true;
          break;
        }
        --iter;
        red.erase(iter);
      }
    }
    if (fail) {
      printf("NO\n");
    } else {
      printf("YES\n");
    }
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int tests;
  scanf("%d", &tests);
  for (int test = 0; test < (int)tests; test++) {
    int n;
    scanf("%d", &n);
    vector<int> a(n);
    for (int i = 0; i < (int)n; i++) {
      scanf("%d", &a[i]);
    }
    string s;
    cin >> s;
    multiset<int> red, blue;
    for (int i = 0; i < (int)n; i++) {
      if (s[i] == 'B') {
        blue.insert(a[i]);
      } else {
        red.insert(a[i]);
      }
    }
    bool fail = false;
    for (int i = (int)1; i <= (int)n; i++) {
      while (!blue.empty() && *blue.begin() < i) {
        blue.erase(blue.begin());
      }
      if (!blue.empty()) {
        blue.erase(blue.begin());
      } else {
        auto iter = red.upper_bound(i);
        if (iter == red.begin()) {
          fail = true;
          break;
        }
        --iter;
        red.erase(iter);
      }
    }
    if (fail) {
      printf("NO\n");
    } else {
      printf("YES\n");
    }
  }
}

```