In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.

The i-th conveyor belt is in one of three states:

  * If it is clockwise, snakes can only go from room i to (i+1) mod n. 
  * If it is anticlockwise, snakes can only go from room (i+1) mod n to i. 
  * If it is off, snakes can travel in either direction. 

<image>

Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.

Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?

Input

Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows. 

The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.

The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.

  * If s_{i} =  '>', the i-th conveyor belt goes clockwise. 
  * If s_{i} =  '<', the i-th conveyor belt goes anticlockwise. 
  * If s_{i} =  '-', the i-th conveyor belt is off. 



It is guaranteed that the sum of n among all test cases does not exceed 300 000.

Output

For each test case, output the number of returnable rooms.

Example

Input


4
4
-&gt;&lt;-
5
&gt;&gt;&gt;&gt;&gt;
3
&lt;--
2
&lt;&gt;


Output


3
5
3
0

Note

In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.

In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
import java.util.*;
import java.io.*;

public class Belted_Rooms {

	static int mod = (int) (1e9 + 7);

	public static void main(String[] args) throws java.lang.Exception {
		long start = System.nanoTime(); // Program Start
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {
			int n=Integer.parseInt(br.readLine());
			String c=br.readLine();
			c+=c.charAt(0);
			char s[]=c.toCharArray();
			boolean hasCW=false,hasACW=false;
			for(int i=0; i<n; i++) {
				if(s[i]=='>') hasCW=true;
				if(s[i]=='<') hasACW=true;
			}
			if(hasCW && hasACW) {
				int ans=0;
				for(int i=0; i<n; i++) {
					if(s[i]=='-' || s[i+1]=='-') ++ans;
				}
				System.out.println(ans);
			}else {
				System.out.println(n);
			}
			
		}
		long end = System.nanoTime(); // Program End
		System.err.println("Time taken: " + (end - start) / 1000000 + " ms");
	}

	static class CP {

		static boolean isPrime(long n) {
			if (n <= 1)
				return false;
			if (n == 2 || n == 3)
				return true;
			if (n % 2 == 0 || n % 3 == 0)
				return false;
			for (int i = 5; i * i <= n; i += 6) {
				if (n % i == 0 || n % (i + 2) == 0)
					return false;
			}
			return true;
		}

		static long binary_Expo(long a, long b) { // calculating a^b
			long res = 1;
			while (b != 0) {
				if ((b & 1) == 1) {
					res *= a;
					--b;
				}
				a *= a;
				b /= 2;
			}
			return res;
		}

		static long Modular_Expo(long a, long b) {
			long res = 1;
			while (b != 0) {
				if ((b & 1) == 1) {
					res = (res * a) % mod;
					--b;
				}
				a = (a * a) % mod;
				b /= 2;
			}
			return res;
		}

		static int i_gcd(int a, int b) {// iterative way to calculate gcd.
			while (true) {
				if (b == 0)
					return a;
				int c = a;
				a = b;
				b = c % b;
			}
		}

		static long gcd(long a, long b) {// here b is the remainder
			if (b == 0)
				return a; //because each time b will divide a.
			return gcd(b, a % b);
		}

		static long ceil_div(long a, long b) { // a numerator b denominator
			return (a + b - 1) / b;
		}

		static int getIthBitFromInt(int bits, int i) {
			return (bits >> (i - 1)) & 1;
		}

		static int upper_Bound(int a[], int x) {//closest to the left+1
			int l = -1, r = a.length;
			while (l + 1 < r) {
				int m = (l + r) >>> 1;
				if (a[m] <= x)
					l = m;
				else
					r = m;
			}
			return l + 1;
		}

		static int lower_Bound(int a[], int x) {//closest to the right
			int l = -1, r = a.length;
			while (l + 1 < r) {
				int m = (l + r) >>> 1;
				if (a[m] >= x)
					r = m;
				else
					l = m;
			}
			return r;
		}

		static void sort(int a[]) {//heap sort
			PriorityQueue<Integer> q = new PriorityQueue<>();
			for (int i = 0; i < a.length; i++)
				q.add(a[i]);
			for (int i = 0; i < a.length; i++)
				a[i] = q.poll();
		}

		static void countSort(int arr[], int n, int exp) {
			int output[] = new int[n]; // output array
			int i;
			int count[] = new int[10];
			Arrays.fill(count, 0);

			// Store count of occurrences in count[]
			for (i = 0; i < n; i++)
				count[(arr[i] / exp) % 10]++;

			// Change count[i] so that count[i] now contains
			// actual position of this digit in output[]
			for (i = 1; i < 10; i++)
				count[i] += count[i - 1];

			// Build the output array
			for (i = n - 1; i >= 0; i--) {
				output[count[(arr[i] / exp) % 10] - 1] = arr[i];
				count[(arr[i] / exp) % 10]--;
			}

			// Copy the output array to arr[], so that arr[] now
			// contains sorted numbers according to curent digit
			for (i = 0; i < n; i++)
				arr[i] = output[i];
		}

		// The main function to that sorts arr[] of size n using
		// Radix Sort
		static void radixsort(int arr[], int n) {
			// Find the maximum number to know number of digits
			int m = getMax(arr, n);

			// Do counting sort for every digit. Note that
			// instead of passing digit number, exp is passed.
			// exp is 10^i where i is current digit number
			for (int exp = 1; m / exp > 0; exp *= 10)
				countSort(arr, n, exp);
		}

		static int getMax(int arr[], int n) {
			int mx = arr[0];
			for (int i = 1; i < n; i++)
				if (arr[i] > mx)
					mx = arr[i];
			return mx;
		}
	}
}