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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-12;
const int N = 33000;
const int M = 1007;
vector<pair<int, int> > ev[N * 4];
void insert(int left, int right, int l, int r, int x, pair<int, int> v) {
  if (l <= left && r >= right) {
    ev[x].push_back(v);
    return;
  }
  int mid = (left + right) >> 1;
  if (l <= mid) {
    insert(left, mid, l, r, x * 2, v);
  }
  if (r > mid) {
    insert(mid + 1, right, l, r, x * 2 + 1, v);
  }
}
int dp[M];
int rec[30][M];
vector<int> query[N];
int k;
const int p = 10000000 + 19;
const int q = 1000000007;
int ans[N];
void go(pair<int, int> x) {
  for (int i = k; i >= x.first; --i) {
    dp[i] = max(dp[i], dp[i - x.first] + x.second);
  }
}
void dfs(int left, int right, int x, int level) {
  for (int i = 0; i < k + 1; i++) rec[level][i] = dp[i];
  for (__typeof(ev[x].begin()) it = ev[x].begin(); it != ev[x].end(); it++) {
    go(*it);
  }
  if (left != right) {
    int mid = (left + right) >> 1;
    dfs(left, mid, x * 2, level + 1);
    dfs(mid + 1, right, x * 2 + 1, level + 1);
  } else {
    for (__typeof(query[left].begin()) it = query[left].begin();
         it != query[left].end(); it++) {
      long long coef = 1;
      int hashes = 0;
      for (int i = 1; i <= k; ++i) {
        hashes = (hashes + dp[i] * coef) % q;
        coef = coef * p % q;
      }
      cout << hashes << endl;
    }
  }
  for (int i = 0; i < k + 1; i++) dp[i] = rec[level][i];
}
int main() {
  int n, q;
  cin >> n >> k;
  vector<pair<int, int> > g(n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &g[i].second, &g[i].first);
  }
  vector<int> when(n);
  fill(when.begin(), when.end(), 0);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int v, w;
      scanf("%d%d", &v, &w);
      when.push_back(i + 1);
      g.push_back(make_pair(w, v));
    } else if (tp == 2) {
      int which;
      scanf("%d", &which);
      --which;
      insert(0, q, when[which], i + 1, 1, g[which]);
      when[which] = -1;
    } else if (tp == 3) {
      query[i + 1].push_back(i);
    }
  }
  for (int i = 0; i < when.size(); i++) {
    if (when[i] != -1) {
      insert(0, q, when[i], q, 1, g[i]);
    }
  }
  dfs(0, q, 1, 0);
  return 0;
}