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 int P = 1e7 + 19, MOD = 1e9 + 7;
const int MAXN = 5001;
const int MAXQ = 3e4 + 1;
bool query_leaf[1 << 16];
int n, k, q;
int v[MAXQ + MAXN], m[MAXQ + MAXN], lo[MAXQ + MAXN], hi[MAXQ + MAXN];
inline int ADD(int a, int b) {
  a += b;
  while (a < 0) a += MOD;
  return a % MOD;
}
inline int MUL(int a, int b) { return (long long)a * b % MOD; }
class Knapsack {
 private:
  vector<int> best;

 public:
  Knapsack() { best.resize(k + 1); }
  void add(vector<pair<int, int>> &items) {
    int v, m;
    for (auto x : items) {
      tie(v, m) = x;
      for (int j = k; j > 0; --j)
        if (j - m >= 0) best[j] = max(best[j], best[j - m] + v);
    }
  }
  int get_score() {
    int ret = 0, p = 1;
    for (int i = 1; i <= k; ++i) {
      ret = ADD(ret, MUL(p, best[i]));
      p = MUL(p, P);
    }
    return ret;
  }
};
stack<Knapsack> S;
class Tournament {
 private:
  int offset, lo, hi;
  vector<vector<pair<int, int>>> tree;
  void insert(int node, int from, int to, pair<int, int> &item) {
    if (hi <= from || to <= lo) return;
    if (from >= lo && to <= hi) {
      tree[node].push_back(item);
      return;
    }
    insert(2 * node, from, (from + to) / 2, item);
    insert(2 * node + 1, (from + to) / 2, to, item);
  }
  void solve(int node, int from, int to) {
    Knapsack curr = S.top();
    curr.add(tree[node]);
    S.push(curr);
    if (from + 1 == to) {
      if (query_leaf[from]) printf("%d\n", S.top().get_score());
      S.pop();
      return;
    }
    solve(2 * node, from, (from + to) / 2);
    solve(2 * node + 1, (from + to) / 2, to);
    S.pop();
  }

 public:
  Tournament() {}
  void init(int n) {
    for (offset = 1; offset <= n; offset <<= 1)
      ;
    tree.resize(2 * offset);
  }
  void insert(pair<int, int> item, int _lo, int _hi) {
    lo = _lo;
    hi = _hi;
    insert(1, 0, offset, item);
  }
  void dfs() { solve(1, 0, offset); }
};
int main(void) {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d%d", &v[i], &m[i]);
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &v[n], &m[n]);
      lo[n++] = i;
    }
    if (type == 2) {
      int x;
      scanf("%d", &x);
      --x;
      hi[x] = i + 1;
    }
    if (type == 3) query_leaf[i] = true;
  }
  Tournament T;
  T.init(q);
  for (int i = 0; i < n; ++i) {
    if (hi[i] == 0) hi[i] = q + 1;
    --hi[i];
    if (lo[i] < hi[i]) T.insert({v[i], m[i]}, lo[i], hi[i]);
  }
  S.push(Knapsack());
  T.dfs();
  return 0;
}