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;
const long long cp = 1e7 + 19;
const long long mod = 1e9 + 7;
int add[20202];
int rm[20202];
long long v[20202];
int w[20202];
long long vv[30303];
int qq[30303];
int n, k;
long long dp[1010];
void addn(int e, vector<pair<int, long long> >& st) {
  for (int i = k; i >= w[e]; i--) {
    if (dp[i - w[e]] != -1) {
      if (dp[i] < dp[i - w[e]] + v[e]) {
        st.push_back({i, dp[i]});
        dp[i] = dp[i - w[e]] + v[e];
      }
    }
  }
}
void go(int l, int r, vector<int> es) {
  if (l > r) return;
  vector<pair<int, long long> > st;
  vector<int> ces;
  for (int e : es) {
    if (rm[e] < l || add[e] > r) continue;
    if (add[e] <= l && r <= rm[e]) {
      addn(e, st);
    } else {
      ces.push_back(e);
    }
  }
  if (l == r && qq[l] == 1) {
    long long ma = 0;
    long long pp = 1;
    for (int i = 1; i <= k; i++) {
      ma = max(ma, dp[i]);
      vv[l] += (ma % mod) * (pp % mod);
      vv[l] %= mod;
      pp *= cp;
      pp %= mod;
    }
  }
  if (l < r) {
    int mid = (l + r) / 2;
    go(l, mid, ces);
    go(mid + 1, r, ces);
  }
  while ((int)st.size() > 0) {
    dp[st.back().first] = st.back().second;
    st.pop_back();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= k; i++) {
    dp[i] = -1;
  }
  vector<int> es;
  for (int i = 1; i <= n; i++) {
    cin >> v[i] >> w[i];
    add[i] = 0;
    es.push_back(i);
  }
  int i2 = n + 1;
  int q;
  cin >> q;
  vector<int> q3s;
  for (int i = 1; i <= q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      es.push_back(i2);
      cin >> v[i2] >> w[i2];
      add[i2] = i;
      i2++;
    }
    if (t == 2) {
      int x;
      cin >> x;
      rm[x] = i;
    }
    if (t == 3) {
      q3s.push_back(i);
      qq[i] = 1;
    }
  }
  for (int i = 1; i < i2; i++) {
    if (rm[i] == 0) {
      rm[i] = q + 1;
    }
  }
  go(0, q + 1, es);
  for (int q3 : q3s) {
    cout << vv[q3] << '\n';
  }
}
