Problem: In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array.

A sum of medians of a sorted k-element set S = {a1, a2, ..., ak}, where a1 < a2 < a3 < ... < ak, will be understood by as 

<image>

The <image> operator stands for taking the remainder, that is <image> stands for the remainder of dividing x by y.

To organize exercise testing quickly calculating the sum of medians for a changing set was needed.

Input

The first line contains number n (1 ≤ n ≤ 105), the number of operations performed.

Then each of n lines contains the description of one of the three operations: 

  * add x — add the element x to the set; 
  * del x — delete the element x from the set; 
  * sum — find the sum of medians of the set. 



For any add x operation it is true that the element x is not included in the set directly before the operation.

For any del x operation it is true that the element x is included in the set directly before the operation.

All the numbers in the input are positive integers, not exceeding 109.

Output

For each operation sum print on the single line the sum of medians of the current set. If the set is empty, print 0.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams (also you may use the %I64d specificator).

Examples

Input

6
add 4
add 5
add 1
add 2
add 3
sum


Output

3


Input

14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum


Output

5
11
13

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.InputStream;
import java.lang.*;

public class Source {
	public static void main(String[] args) {
		InputReader in = new InputReader(System.in);
		PrintWriter ou = new PrintWriter(System.out);		
		MainCode obj = new MainCode();
		obj.Solve(in, ou);
		ou.close();
	}
	
}

class MainCode {
	private int n, q;
	private int[] k, a, fa;
	private int[] lazy, size;
	private long[][] tree;
	
	public void Solve(InputReader inp, PrintWriter out) {
		q = inp.nextInt();
		k = new int[q];
		a = new int[q];
		for (int i = 0; i < q; ++i) {
			String S = inp.next();
			if (S.charAt(0) == 'a') {
				k[i] = 0;
				a[i] = inp.nextInt();
			}
			else if (S.charAt(0) == 'd') {
				k[i] = 1;
				a[i] = inp.nextInt();
			}
			else k[i] = 2;
		}
		Compress();
		
		tree = new long[2 * n][5];
		lazy = new int[n];
		size = new int[2 * n];
		for (int i = 0; i < q; ++i) {
			if (k[i] == 0) Update(fa[i], n - 1, a[i]); else
			if (k[i] == 1) Update(fa[i], n - 1, -a[i]); else
				out.println(Query());
		}
	}
	
	private void Compress() {
		Integer[] p = new Integer[q];
		for (int i = 0; i < q; ++i) p[i] = i;
		Arrays.sort(p, new Comparator<Integer>() {
			public int compare(Integer x, Integer y) {
				return a[x] - a[y];
			}
		});
		
		fa = new int[q]; fa[p[0]] = (a[p[0]] == 0) ? -1 : 0;
		for (int i = 1; i < q; ++i) {
			fa[p[i]] = fa[p[i - 1]];
			if (a[p[i]] != a[p[i - 1]]) fa[p[i]]++;
		}
		n = fa[p[q - 1]] + 1;
	}
	
	void Rotate(int k, int v) {
		long[] temp = new long[5];
		for (int i = 0; i < 5; ++i) temp[i] = tree[k][(i + 5 - v) % 5];
		for (int i = 0; i < 5; ++i) tree[k][i] = temp[i];
	}
	
	void Apply(int k, int v) {
		if (k < n) lazy[k] = (lazy[k] + v) % 5;
		Rotate(k, v);
	}
	
	void PushUp(int k) {
		while (k > 1) {
			k >>= 1;
			size[k] = size[k << 1] + size[k << 1 | 1];
			for (int i = 0; i < 5; ++i) tree[k][i] = tree[k << 1][i] + tree[k << 1 | 1][i];
			Rotate(k, lazy[k]);
		}
	}
	
	void PushDown(int k) {
		for (int i = 20; i > 0; --i) {
			int r = k >> i;
			if (lazy[r] != 0) {
				Apply(r << 1 | 0, lazy[r]);
				Apply(r << 1 | 1, lazy[r]);
				lazy[r] = 0;
			}
		}
	}
	
	void Add(int i, int v) {
		size[i + n] += (v > 0) ? 1 : -1;
		tree[i + n][Count(0, i - 1) % 5] += v;
		PushUp(i + n);
	}
	
	void Update(int l, int r, int v) {
		l += n; r += n;
		int lCopy = l;
		int rCopy = r;
		PushDown(l);
		PushDown(r);
		if (v < 0) Add(lCopy - n, v);
		int k = (v > 0) ? 1 : 4;
		for (; l <= r; l = (l + 1) >> 1, r = (r - 1) >> 1) {
			if ((l & 1) == 1) Apply(l++, k);
			if ((r & 1) == 0) Apply(r--, k);
		}
		PushUp(lCopy);
		PushUp(rCopy);
		if (v > 0) {
			PushDown(lCopy);
			PushDown(rCopy);
			Add(lCopy - n, v);
		}
	}
	
	int Count(int l, int r) {
		int res = 0;
		for (l += n, r += n; l <= r; l = (l + 1) >> 1, r = (r - 1) >> 1) {
			if ((l & 1) == 1) res += size[l++];
			if ((r & 1) == 0) res += size[r--];
		}
		return res;
	}
	
	long Query() {
		return (n == 0) ? 0 : tree[1][2];
	}
	
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }
    
    public long nextLong() {
    	return Long.parseLong(next());
    }
    
    public double nextDouble() {
    	return Double.parseDouble(next());
    }

}