There's a famous museum in the city where Kleofáš lives. In the museum, n exhibits (numbered 1 through n) had been displayed for a long time; the i-th of those exhibits has value vi and mass wi. 

Then, the museum was bought by a large financial group and started to vary the exhibits. At about the same time, Kleofáš... gained interest in the museum, so to say.

You should process q events of three types:

  * type 1 — the museum displays an exhibit with value v and mass w; the exhibit displayed in the i-th event of this type is numbered n + i (see sample explanation for more details)
  * type 2 — the museum removes the exhibit with number x and stores it safely in its vault
  * type 3 — Kleofáš visits the museum and wonders (for no important reason at all, of course): if there was a robbery and exhibits with total mass at most m were stolen, what would their maximum possible total value be?



For each event of type 3, let s(m) be the maximum possible total value of stolen exhibits with total mass  ≤ m.

Formally, let D be the set of numbers of all exhibits that are currently displayed (so initially D = {1, ..., n}). Let P(D) be the set of all subsets of D and let 

<image>

Then, s(m) is defined as 

<image>

Compute s(m) for each <image>. Note that the output follows a special format.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 1000) — the initial number of exhibits in the museum and the maximum interesting mass of stolen exhibits. 

Then, n lines follow. The i-th of them contains two space-separated positive integers vi and wi (1 ≤ vi ≤ 1 000 000, 1 ≤ wi ≤ 1000) — the value and mass of the i-th exhibit.

The next line contains a single integer q (1 ≤ q ≤ 30 000) — the number of events.

Each of the next q lines contains the description of one event in the following format:

  * 1 v w — an event of type 1, a new exhibit with value v and mass w has been added (1 ≤ v ≤ 1 000 000, 1 ≤ w ≤ 1000)
  * 2 x — an event of type 2, the exhibit with number x has been removed; it's guaranteed that the removed exhibit had been displayed at that time
  * 3 — an event of type 3, Kleofáš visits the museum and asks his question



There will be at most 10 000 events of type 1 and at least one event of type 3.

Output

As the number of values s(m) can get large, output the answers to events of type 3 in a special format.

For each event of type 3, consider the values s(m) computed for the question that Kleofáš asked in this event; print one line containing a single number 

<image>

where p = 107 + 19 and q = 109 + 7.

Print the answers to events of type 3 in the order in which they appear in the input.

Examples

Input

3 10
30 4
60 6
5 1
9
3
1 42 5
1 20 3
3
2 2
2 4
3
1 40 6
3


Output

556674384
168191145
947033915
181541912


Input

3 1000
100 42
100 47
400 15
4
2 2
2 1
2 3
3


Output

0

Note

In the first sample, the numbers of displayed exhibits and values s(1), ..., s(10) for individual events of type 3 are, in order: 

<image> <image> <image> <image>

The values of individual exhibits are v1 = 30, v2 = 60, v3 = 5, v4 = 42, v5 = 20, v6 = 40 and their masses are w1 = 4, w2 = 6, w3 = 1, w4 = 5, w5 = 3, w6 = 6.

In the second sample, the only question is asked after removing all exhibits, so s(m) = 0 for any m.
import java.util.*;
import java.io.*;
public class AMuseumRobbery
{
	public static final long MOD = 1000000007;
	public static final long P = 10000019;
	
	public static PrintWriter out;
	public static int[] values;
	public static int[] weights;
	public static ArrayList<Integer>[] tree;
	public static int[][] dp;
	public static int state;
	
	public static void main(String[] args) throws Exception
	{
		FastScanner in = new FastScanner(System.in);
		out = new PrintWriter(System.out);
		
		int n = in.nextInt();
		int k = in.nextInt();
		
		values = new int[n + 10000];
		weights = new int[n + 10000];
		int[] left = new int[n + 10000];
		int[] right = new int[n + 10000];
		Arrays.fill(right, -2);
		for(int x = 0; x < n; x++)
		{
			values[x] = in.nextInt();
			weights[x] = in.nextInt();
			left[x] = 0;
		}
		
		int q = in.nextInt();
		
		int pos = 0;
		int index = n;
		for(int y = 0; y < q; y++)
		{
			int type = in.nextInt();
			if(type == 1)
			{
				values[index] = in.nextInt();
				weights[index] = in.nextInt();
				left[index] = pos;
				
				index++;
			}
			else if(type == 2)
			{
				int x = in.nextInt() - 1;
				
				right[x] = pos - 1;
			}
			else
			{
				pos++;
			}
		}
		
		for(int z = 0; z < index; z++)
		{
			if(right[z] == -2)
			{
				right[z] = pos - 1;
			}
		}
		
		tree = new ArrayList[pos << 2 | 1];
		for(int a = 0; a < tree.length; a++)
		{
			tree[a] = new ArrayList<Integer>(2);
		}
		
		for(int b = 0; b < index; b++)
		{
			add(1, 0, pos - 1, left[b], right[b], b);
		}
		
		dp = new int[index + 1][k + 1];
		state = 1;
		
		calc(1, 0, pos - 1);
		
		out.close();
	}
	
	public static void add(int i, int l, int r, int left, int right, int val)
	{
		if(r < left || l > right)
		{
			return;
		}
		else if(l >= left && r <= right)
		{
			tree[i].add(val);
		}
		else
		{
			int mid = (l + r) >> 1;
			add(i << 1, l, mid, left, right, val);
			add(i << 1 | 1, mid + 1, r, left, right, val);
		}
	}
	
	public static void calc(int i, int l, int r)
	{
		for(int a : tree[i])
		{
			dp[state] = Arrays.copyOf(dp[state - 1], dp[0].length);
			for(int b = 0; b + weights[a] < dp[0].length; b++)
			{
				dp[state][b + weights[a]] = Math.max(dp[state][b + weights[a]], dp[state - 1][b] + values[a]);
			}
			
			state++;
		}
		
		if(l == r)
		{
			long p = 1;
			long result = 0;
			for(int c = 1; c < dp[0].length; c++)
			{
				result += (dp[state - 1][c] % MOD) * p;
				result %= MOD;
				
				p *= P;
				p %= MOD;
			}
			
			out.println(result);
		}
		else
		{
			int mid = (l + r) >> 1;
			calc(i << 1, l, mid);
			calc(i << 1 | 1, mid + 1, r);
		}
		
		state -= tree[i].size();
	}
	
	static class FastScanner
	{
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner(InputStream input)
		{
			br = new BufferedReader(new InputStreamReader(input));
			st = new StringTokenizer("");
		}
		
		public String next() throws IOException
		{
			if(st.hasMoreTokens())
			{
				return st.nextToken();
			}
			else
			{
				st = new StringTokenizer(br.readLine());
				return next();
			}
		}
		
		public int nextInt() throws IOException
		{
			return Integer.parseInt(next());
		}
	}
}
