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;
template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
const int N = 15e3 + 7;
const long long P = 1e7 + 19;
const int Q = 1e9 + 7;
int n, k, in[N], out[N], w[N], v[N];
void update(vector<int> &vec, int id) {
  for (int i = k; i >= w[id]; i--) {
    vec[i] = max(vec[i], vec[i - w[id]] + v[id]);
  }
}
void solve(int l, int r, vector<int> vec, vector<int> &can) {
  if (l == r) {
    for (auto u : can) {
      if (in[u] <= l && r <= out[u]) {
        update(vec, u);
      }
    }
    long long ans = 0;
    for (int m = k; m; m--) {
      ans = (P * ans + vec[m]) % Q;
    }
    cout << ans << "\n";
    return;
  }
  int m = (l + r) / 2;
  vector<int> can2;
  for (auto u : can) {
    if (in[u] <= l && r <= out[u]) {
      update(vec, u);
    } else {
      can2.push_back(u);
    }
  }
  solve(l, m, vec, can2);
  solve(m + 1, r, vec, can2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  int last = 0;
  for (int i = 1; i <= n; i++) {
    cin >> v[i] >> w[i];
    in[i] = last + 1;
    out[i] = 3 * N;
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 3) {
      last++;
    }
    if (t == 2) {
      int x;
      cin >> x;
      out[x] = last;
    }
    if (t == 1) {
      n++;
      cin >> v[n] >> w[n];
      in[n] = last + 1;
      out[n] = 3 * N;
    }
  }
  vector<int> emp(k + 1, 0), can;
  for (int i = 1; i <= n; i++) {
    can.push_back(i);
  }
  solve(1, last, emp, can);
  return 0;
}
