Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class E {
	FastScanner in;
	PrintWriter out;

	int n, k, q;
	int[] t;
	int[] v;
	int[] w;
	int[] id;
	int[] tin;
	int[] tout;
	long[] ans;
	long[] z;
	final long P = (long) (1e7 + 19);
	final long Q = (long) (1e9 + 7);
	
	void sol(int l, int r) {
		if (r - l == 1) {
			if (t[l] == 3) {
				for (int i = k; i >= 1; i--) {
					ans[l] = (ans[l] * P + z[i]) % Q;
				}
			}
			return;
		}
		long[] mem = new long[k + 1];
		for (int i = 0; i <= k; i++) {
			mem[i] = z[i];
		}
		int mid = (l + r) / 2;
		for (int i = l; i < r; i++) {
			if (t[i] != 3 && tin[id[i]] < l && tout[id[i]] >= mid) {
				for (int j = k - w[id[i]]; j >= 0; j--) {
					z[j + w[id[i]]] = Math.max(z[j + w[id[i]]], z[j] + v[id[i]]);
				}
			}
		}
		sol(l, mid);
		
		for (int i = 0; i <= k; i++) {
			z[i] = mem[i];
		}
		for (int i = l; i < r; i++) {
			if (t[i] != 3 && tin[id[i]] < mid && tout[id[i]] >= r) {
				for (int j = k - w[id[i]]; j >= 0; j--) {
					z[j + w[id[i]]] = Math.max(z[j + w[id[i]]], z[j] + v[id[i]]);
				}
			}
		}
		sol(mid, r);
	}
	
	public void solve() throws IOException {
		n = in.nextInt();
		k = in.nextInt();
		int[] initV = new int[n];
		int[] initW = new int[n];
		for (int i = 0; i < n; i++) {
			initV[i] = in.nextInt();
			initW[i] = in.nextInt();
		}
		q = in.nextInt();
		q += n;
		t = new int[q];
		v = new int[q];
		w = new int[q];
		id = new int[q];
		ans = new long[q];
		tin = new int[q];
		tout = new int[q];
		Arrays.fill(tout, q);
		z = new long[k + 1];
		for (int i = 0; i < n; i++) {
			t[i] = 1;
			id[i] = i;
			v[i] = initV[i];
			w[i] = initW[i];
			tin[id[i]] = i;
		}
		//System.err.println(q);
		int addId = n;
		for (int i = n; i < q; i++) {
			//System.err.println("wait " + i);
			t[i] = in.nextInt();
			//System.err.println(t[i]);
			if (t[i] == 1) {
				id[i] = addId++;
				v[id[i]] = in.nextInt();
				w[id[i]] = in.nextInt();
				tin[id[i]] = i;
			} else if (t[i] == 2) {
				id[i] = in.nextInt() - 1;
				tout[id[i]] = i;
			}
		}
		//System.out.println("start");
		sol(0, q);
		for (int i = 0; i < q; i++) {
			if (t[i] == 3) {
				out.println(ans[i]);
			}
		}
	}

	public void run() {
		try {
			in = new FastScanner();
			out = new PrintWriter(System.out);

			solve();

			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

	public static void main(String[] arg) {
		new E().run();
	}
}