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 oo = 0x3f3f3f3f;
const double eps = 1e-9;
const int maxn = 30000 + 100;
const long long PP = 10000019;
const long long mod = 1000000007;
int B[maxn], E[maxn], V[maxn], W[maxn];
vector<int> st[4 * maxn];
void push(int p, int b, int e, int x, int y, int v) {
  if (x <= b && e <= y)
    st[p].push_back(v);
  else {
    int m = (b + e) >> 1, l = p << 1, r = l | 1;
    if (x < m) push(l, b, m, x, y, v);
    if (m < y) push(r, m, e, x, y, v);
  }
}
vector<int> question;
int current = 0;
int dp[20][maxn], n, k;
void solve(int p, int b, int e, int lv) {
  for (int i = 0; i <= k; ++i) dp[lv][i] = dp[lv - 1][i];
  for (auto idx : st[p]) {
    int v = V[idx], w = W[idx];
    for (int i = k; i >= w; --i) dp[lv][i] = max(dp[lv][i], dp[lv][i - w] + v);
  }
  if (b + 1 == e) {
    if (question[current] == b) {
      current++;
      long long answer = 0;
      for (int i = k; i; --i) answer = (answer * PP % mod + dp[lv][i]) % mod;
      cout << answer << '\n';
    }
  } else {
    int m = (b + e) >> 1, l = p << 1, r = l | 1;
    solve(l, b, m, lv + 1);
    solve(r, m, e, lv + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    B[i] = 0, E[i] = oo;
    cin >> V[i] >> W[i];
  }
  int t = n;
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    int typ;
    cin >> typ;
    if (typ == 1) {
      t++;
      B[t] = i, E[t] = oo;
      cin >> V[t] >> W[t];
    } else if (typ == 2) {
      int u;
      cin >> u;
      E[u] = i + 1;
    } else {
      question.push_back(i);
    }
  }
  question.push_back(oo);
  q++;
  for (int i = 1; i <= t; ++i) {
    E[i] = min(E[i], q);
    push(1, 0, q, B[i], E[i], i);
  }
  solve(1, 0, q, 1);
  return 0;
}
