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 int MOD = 1e9 + 7;
const int OFF = (1 << 15);
int n, k, q;
pair<pair<int, int>, pair<int, int> > p[15002];
int wh[30002];
int q1[30002];
int q2[30002];
long long pot[1002];
vector<int> v[2 * OFF + 5];
long long dp[OFF + 5][1002];
long long tdp[1002];
void update(int a, int b, int l, int r, int x, int kol) {
  if (r <= a || b <= l) return;
  if (a <= l && r <= b) {
    v[x].push_back(kol);
    return;
  }
  update(a, b, l, (l + r) / 2, x * 2, kol);
  update(a, b, (l + r) / 2, r, x * 2 + 1, kol);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  pot[0] = 1;
  for (int i = 1; i <= k; i++) {
    pot[i] = pot[i - 1] * (1e7 + 19);
    pot[i] %= MOD;
  }
  for (int i = 0; i < n; i++) {
    cin >> p[i].first.first >> p[i].first.second;
    p[i].second.first = 0;
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> wh[i];
    if (wh[i] == 1) {
      cin >> q1[i] >> q2[i];
      p[n] = {{q1[i], q2[i]}, {i, 0}};
      n++;
    } else if (wh[i] == 2) {
      cin >> q1[i];
      q1[i]--;
      p[q1[i]].second.second = i;
    }
  }
  for (int i = 0; i < n; i++) {
    if (p[i].second.second == 0) p[i].second.second = q + 1;
    update(p[i].second.first, p[i].second.second + 1, 0, OFF, 1, i);
  }
  for (int i = 1; i < OFF; i++) {
    for (int l = 0; l < v[i].size(); l++) {
      for (int j = k; j >= 0; j--) {
        int t = v[i][l];
        if (j - p[t].first.second >= 0) {
          dp[i][j] =
              max(dp[i][j], dp[i][j - p[t].first.second] + p[t].first.first);
        }
        if (j > 0) dp[i][j] = max(dp[i][j], dp[i][j - 1]);
      }
    }
    for (int j = 0; j <= k; j++) {
      if (i * 2 < OFF) dp[i * 2][j] = dp[i][j];
      if (i * 2 + 1 < OFF) dp[i * 2 + 1][j] = dp[i][j];
    }
  }
  for (int i = 1; i <= q; i++) {
    if (wh[i] == 3) {
      long long sol = 0;
      int c = OFF + i;
      c /= 2;
      for (int j = 1; j <= k; j++) tdp[j] = dp[c][j];
      for (int l = 0; l < v[OFF + i].size(); l++) {
        int t = v[OFF + i][l];
        for (int j = 1; j <= k; j++) {
          if (j - p[t].first.second >= 0) {
            tdp[j] = max(tdp[j], tdp[j - p[t].first.second] + p[t].first.first);
          }
          if (j > 0) tdp[j] = max(tdp[j], tdp[j - 1]);
        }
      }
      for (int j = 1; j <= k; j++) {
        sol += (tdp[j] % MOD * pot[j - 1]) % MOD;
        sol %= MOD;
      }
      cout << sol << endl;
    }
  }
  return 0;
}
