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.
#include <bits/stdc++.h>
using namespace std;
int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 5000, Q = 30000, QQ = 32768, BASE = 1e7 + 19, MOD = 1e9 + 7;
int k, v[N + Q], w[N + Q], s[N + Q], e[N + Q];
int op[Q], ans[QQ];
vector<int> seg[QQ * 2];
void add(int l, int r, int x) {
  for (l += QQ, r += QQ; l < r; l >>= 1, r >>= 1) {
    if (l & 1) seg[l++].push_back(x);
    if (r & 1) seg[--r].push_back(x);
  }
}
void topdown(int rt, vector<int> dp) {
  if (rt >= QQ * 2) return;
  for (int i : seg[rt])
    for (int j = (k + 1); --j >= (w[i]);)
      dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
  if (rt < QQ) {
    topdown(rt * 2, dp);
    topdown(rt * 2 + 1, dp);
  } else {
    int d = 0;
    long long p = 1;
    for (int i = 1; i <= (k); i++) {
      d = max(d, dp[i]);
      ans[rt - QQ] = (ans[rt - QQ] + p * d) % MOD;
      p = (p * BASE) % MOD;
    }
  }
}
int main() {
  int n = ri();
  k = ri();
  for (int i = 0; i < (n); i++) {
    v[i] = ri();
    w[i] = ri();
  }
  int q = ri(), tick = 0;
  fill_n(e, n + q, -1);
  for (int i = 0; i < (q); i++) {
    int op = ri();
    if (op == 1) {
      v[n] = ri();
      w[n] = ri();
      s[n++] = tick;
    } else if (op == 2)
      e[ri() - 1] = tick;
    else
      tick++;
  }
  for (int i = 0; i < (n); i++) add(s[i], ~e[i] ? e[i] : tick, i);
  topdown(1, vector<int>(k + 1));
  for (int i = 0; i < (tick); i++) printf("%d\n", ans[i]);
}
