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;
int vl[300005], wt[300005];
vector<int> va[300005], vr[300005];
bool use[300005];
int n, k, q;
vector<vector<int> > solve(vector<vector<int> > add, vector<vector<int> > rem) {
  if (add.size() == 1) {
    vector<int> dp;
    dp.resize(k + 1, 0);
    for (int i = 0; i < add[0].size(); i++) {
      int v = add[0][i];
      for (int j = k - wt[v]; j >= 0; j--) {
        dp[j + wt[v]] = max(dp[j + wt[v]], dp[j] + vl[v]);
      }
    }
    vector<vector<int> > ret;
    ret.push_back(dp);
    return ret;
  }
  vector<vector<int> > na, nr;
  for (int i = 0; i < add.size(); i++) {
    va[i].clear();
    vr[i].clear();
    vector<int> A = add[i], R = rem[i];
    for (int j = 0; j < R.size(); j++) use[R[j]] = true;
    for (int j = 0; j < A.size(); j++) {
      if (!use[A[j]])
        va[i].push_back(A[j]);
      else
        use[A[j]] = false;
    }
    for (int j = 0; j < R.size(); j++) {
      if (use[R[j]]) {
        vr[i].push_back(R[j]);
        use[R[j]] = false;
      }
    }
  }
  for (int i = 0; i < add.size(); i += 2) {
    vector<int> A, R;
    if (i + 1 < add.size()) {
      if (i - 1 >= 0) {
        for (int j = 0; j < rem[i].size(); j++) use[rem[i][j]] = true;
        for (int j = 0; j < va[i - 1].size(); j++)
          if (!use[va[i - 1][j]]) A.push_back(va[i - 1][j]);
        for (int j = 0; j < rem[i].size(); j++) use[rem[i][j]] = false;
      }
      if (i + 3 < add.size()) {
        for (int j = 0; j < add[i + 1].size(); j++) use[add[i + 1][j]] = true;
        for (int j = 0; j < vr[i + 2].size(); j++)
          if (!use[vr[i + 2][j]]) R.push_back(vr[i + 2][j]);
        for (int j = 0; j < add[i + 1].size(); j++) use[add[i + 1][j]] = false;
      }
      for (int j = 0; j < va[i].size(); j++) A.push_back(va[i][j]);
      for (int j = 0; j < vr[i + 1].size(); j++) R.push_back(vr[i + 1][j]);
    } else {
      if (i - 1 >= 0) A = va[i - 1];
      for (int j = 0; j < add[i].size(); j++) A.push_back(add[i][j]);
      for (int j = 0; j < rem[i].size(); j++) R.push_back(rem[i][j]);
    }
    na.push_back(A);
    nr.push_back(R);
  }
  vector<vector<int> > dp = solve(na, nr);
  vector<vector<int> > ret;
  for (int i = 0; i < add.size(); i++) {
    va[i].clear();
    vr[i].clear();
    vector<int> A = add[i], R = rem[i];
    for (int j = 0; j < R.size(); j++) use[R[j]] = true;
    for (int j = 0; j < A.size(); j++) {
      if (!use[A[j]])
        va[i].push_back(A[j]);
      else
        use[A[j]] = false;
    }
    for (int j = 0; j < R.size(); j++) {
      if (use[R[j]]) {
        vr[i].push_back(R[j]);
        use[R[j]] = false;
      }
    }
  }
  for (int i = 0; i < add.size(); i++) {
    vector<int> now = dp[i / 2];
    if (i % 2 == 0 && i + 1 < add.size()) {
      for (int j = 0; j < va[i].size(); j++) use[va[i][j]] = true;
      for (int j = 0; j < add[i].size(); j++) {
        if (!use[add[i][j]]) {
          int v = add[i][j];
          for (int t = k - wt[v]; t >= 0; t--) {
            now[t + wt[v]] = max(now[t + wt[v]], now[t] + vl[v]);
          }
        }
      }
      for (int j = 0; j < va[i].size(); j++) use[va[i][j]] = false;
      if (i > 0) {
        for (int j = 0; j < vr[i].size(); j++) {
          int v = vr[i][j];
          for (int t = k - wt[v]; t >= 0; t--) {
            now[t + wt[v]] = max(now[t + wt[v]], now[t] + vl[v]);
          }
        }
      }
    } else if (i % 2 == 1) {
      for (int j = 0; j < add[i].size(); j++) {
        int v = add[i][j];
        for (int t = k - wt[v]; t >= 0; t--) {
          now[t + wt[v]] = max(now[t + wt[v]], now[t] + vl[v]);
        }
      }
    }
    ret.push_back(now);
  }
  return ret;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d %d", &vl[i], &wt[i]);
  vector<vector<int> > add, rem;
  vector<int> A, R;
  for (int i = 0; i < n; i++) A.push_back(i);
  int sz = n;
  scanf("%d", &q);
  memset(use, false, sizeof(use));
  for (int i = 0; i < q; i++) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d %d", &vl[sz], &wt[sz]);
      A.push_back(sz);
      sz++;
    } else if (tp == 2) {
      int x;
      scanf("%d", &x);
      x--;
      R.push_back(x);
    } else {
      vector<int> na, nr;
      for (int j = 0; j < R.size(); j++) use[R[j]] = true;
      for (int j = 0; j < A.size(); j++) {
        if (!use[A[j]])
          na.push_back(A[j]);
        else
          use[A[j]] = false;
      }
      for (int j = 0; j < R.size(); j++) {
        if (use[R[j]]) {
          nr.push_back(R[j]);
          use[R[j]] = false;
        }
      }
      if (!add.empty()) rem.push_back(nr);
      add.push_back(na);
      A.clear();
      R.clear();
    }
  }
  vector<int> na, nr;
  for (int j = 0; j < R.size(); j++) use[R[j]] = true;
  for (int j = 0; j < A.size(); j++) {
    if (!use[A[j]])
      na.push_back(A[j]);
    else
      use[A[j]] = false;
  }
  for (int j = 0; j < R.size(); j++) {
    if (use[R[j]]) {
      nr.push_back(R[j]);
      use[R[j]] = false;
    }
  }
  rem.push_back(nr);
  vector<vector<int> > dp = solve(add, rem);
  for (int i = 0; i < add.size(); i++) {
    long long int ret = 0, t = 1;
    for (int j = 1; j <= k; j++) {
      ret += (long long int)dp[i][j] * t % 1000000007LL;
      ret %= 1000000007LL;
      t *= 10000019LL;
      t %= 1000000007LL;
    }
    printf("%lld\n", ret);
  }
  return 0;
}