She is an extraordinary girl. She works for a library. Since she is young and cute, she is forced to do a lot of laborious jobs. The most annoying job for her is to put returned books into shelves, carrying them by a cart. The cart is large enough to carry many books, but too heavy for her. Since she is delicate, she wants to walk as short as possible when doing this job. The library has 4N shelves (1 <= N <= 10000), three main passages, and N + 1 sub passages. Each shelf is numbered between 1 to 4N as shown in Figure 1. Horizontal dashed lines correspond to main passages, while vertical dashed lines correspond to sub passages. She starts to return books from the position with white circle in the figure and ends at the position with black circle. For simplicity, assume she can only stop at either the middle of shelves or the intersection of passages. At the middle of shelves, she put books with the same ID as the shelves. For example, when she stops at the middle of shelf 2 and 3, she can put books with ID 2 and 3.

<image>

Since she is so lazy that she doesn’t want to memorize a complicated route, she only walks main passages in forward direction (see an arrow in Figure 1). The walk from an intersection to its adjacent intersections takes 1 cost. It means the walk from the middle of shelves to its adjacent intersection, and vice versa, takes 0.5 cost. You, as only a programmer amoung her friends, are to help her compute the minimum possible cost she takes to put all books in the shelves.



Input

The first line of input contains the number of test cases, T . Then T test cases follow. Each test case consists of two lines. The first line contains the number N , and the second line contains 4N characters, either Y or N . Y in n-th position means there are some books with ID n, and N means no book with ID n.

Output

The output should consists of T lines, each of which contains one integer, the minimum possible cost, for each test case.

Example

Input

2
2
YNNNNYYY
4
NYNNYYNNNYNYYNNN


Output

6
9
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
typedef long long int ll;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

int** dp;
bool** map;
char* buf;

void func(){
	int N,limit;
	scanf("%d",&N);

	for(int i = 0; i < 3; i++){
		for(int k = 0; k <= N; k++)dp[i][k] = BIG_NUM;
	}

	for(int i = 0; i < 2; i++){
		for(int k = 0; k <= 2*N; k++)map[i][k] = false;
	}

	scanf("%s",buf);

	limit = 2*N;

	for(int i = 0; i < limit;i++){
		if(buf[i] == 'Y'){
			map[0][i] = true;
		}
	}
	for(int i = 0; i < limit; i++){
		if(buf[limit+i] == 'Y'){
			map[1][i] = true;
		}
	}

	if(map[0][0] == true && map[1][0] == true){
		dp[0][0] = 3;
		dp[1][0] = 2;
		dp[2][0] = 2;
	}else if(map[0][0] == false && map[1][0] == false){
		dp[0][0] = 0;
		dp[1][0] = 1;
		dp[2][0] = 2;
	}else if(map[0][0] == true && map[1][0] == false){
		dp[0][0] = 1;
		dp[1][0] = 1;
		dp[2][0] = 2;
	}else if(map[0][0] == false && map[1][0] == true){
		dp[0][0] = 3;
		dp[1][0] = 2;
		dp[2][0] = 2;
	}

	limit = N-1;
	for(int i = 1; i <= limit; i++){
		if((map[0][2*i-1] == true || map[0][2*i] == true)&& (map[1][2*i-1] == true || map[1][2*i] == true)){ //??????????£???¨?\\????£???????????????¬????????´???????????????????????´???
			dp[0][i] = min(dp[0][i-1]+4,min(dp[1][i-1]+3,dp[2][i-1]+3));
			dp[1][i] = min(dp[0][i-1]+3,min(dp[1][i-1]+3,dp[2][i-1]+3));
			dp[2][i] = min(dp[0][i-1]+3,min(dp[1][i-1]+3,dp[2][i-1]+4));
		}else if((map[0][2*i-1] == false && map[0][2*i] == false) && (map[1][2*i-1] == false && map[1][2*i] == false)){ //??????????????´???????????????????????´???
			dp[0][i] = min(dp[0][i-1]+1,min(dp[1][i-1]+2,dp[2][i-1]+3));
			dp[1][i] = min(dp[0][i-1]+2,min(dp[1][i-1]+1,dp[2][i-1]+2));
			dp[2][i] = min(dp[0][i-1]+3,min(dp[1][i-1]+2,dp[2][i-1]+1));
		}else if((map[0][2*i-1] == true || map[0][2*i] == true) && (map[1][2*i-1] == false && map[1][2*i] == false)){ //??????????£???????????????´???????????????????????´???
			dp[0][i] = min(dp[0][i-1]+2,min(dp[1][i-1]+2,dp[2][i-1]+3));
			dp[1][i] = min(dp[0][i-1]+2,min(dp[1][i-1]+2,dp[2][i-1]+3));
			dp[2][i] = min(dp[0][i-1]+3,min(dp[1][i-1]+3,dp[2][i-1]+4));
		}else if((map[0][2*i-1] == false && map[0][2*i] == false) && (map[1][2*i-1] == true || map[1][2*i] == true)){ //?\\????£???????????????´???????????????????????´???
			dp[0][i] = min(dp[0][i-1]+4,min(dp[1][i-1]+3,dp[2][i-1]+3));
			dp[1][i] = min(dp[0][i-1]+3,min(dp[1][i-1]+2,dp[2][i-1]+2));
			dp[2][i] = min(dp[0][i-1]+3,min(dp[1][i-1]+2,dp[2][i-1]+2));
		}
	}

	if(map[0][2*N-1] == true && map[1][2*N-1] == true){
		dp[0][N] = min(dp[0][N-1]+4,min(dp[1][N-1]+3,dp[2][N-1]+3));
	}else if(map[0][2*N-1] == false && map[1][2*N-1] == false){
		dp[0][N] = min(dp[0][N-1]+1,min(dp[1][N-1]+2,dp[2][N-1]+3));
	}else if(map[0][2*N-1] == true && map[1][2*N-1] == false){
		dp[0][N] = min(dp[0][N-1]+2,min(dp[1][N-1]+2,dp[2][N-1]+3));
	}else if(map[0][2*N-1] == false && map[1][2*N-1] == true){
		dp[0][N] = min(dp[0][N-1]+4,min(dp[1][N-1]+3,dp[2][N-1]+3));
	}

	printf("%d\n",dp[0][N]);
}

int main(){

	dp = new int*[3];
	for(int i = 0; i < 3; i++){
		dp[i] = new int[10001];
	}

	map = new bool*[2];
	for(int i = 0; i < 2; i++){
		map[i] = new bool[20001];
	}

	buf = new char[40010];


	int TEST;
	scanf("%d",&TEST);

	for(int loop = 0; loop < TEST; loop++){
		func();
	}

	return 0;
}