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>
typedef struct Exhibit {
  int v, w;
  Exhibit(int v, int w) : v(v), w(w) {}
} Exhibit;
std::vector<Exhibit> exh;
std::vector<int> begin, end;
int N, K;
struct DP {
  int dp[1001];
  DP() { std::fill(dp, dp + 1001, 0); }
  void insert(Exhibit e) {
    for (int i = 1000; i >= e.w; i--) {
      dp[i] = std::max(dp[i], dp[i - e.w] + e.v);
    }
  }
};
std::vector<Exhibit> st[30000 * 4];
int qs[30000 * 4];
void insert(int w, int L, int R, int a, int b, Exhibit e) {
  if ((a >= R) || (b <= L)) {
    return;
  }
  if ((a <= L) && (b >= R)) {
    st[w].push_back(e);
    return;
  }
  int M = (L + R) / 2;
  insert(w * 2 + 1, L, M, a, b, e);
  insert(w * 2 + 2, M, R, a, b, e);
}
void query(int w, int L, int R, int i) {
  if ((i >= R) || (i < L)) {
    return;
  }
  qs[w]++;
  if (R - L == 1) {
    return;
  }
  int M = (L + R) / 2;
  query(w * 2 + 1, L, M, i);
  query(w * 2 + 2, M, R, i);
}
std::vector<struct DP> stk;
void dfs(int w, int L, int R) {
  if (qs[w] == 0) {
    return;
  }
  stk.push_back(stk.back());
  for (Exhibit e : st[w]) {
    stk.back().insert(e);
  }
  if (R - L == 1) {
    int64_t sum = 0;
    for (int i = K; i > 0; i--) {
      sum = (sum * 10000019L + stk.back().dp[i]) % 1000000007;
    }
    for (int i = 0; i < qs[w]; i++) {
      std::cout << sum << std::endl;
    }
  } else {
    int M = (L + R) / 2;
    dfs(w * 2 + 1, L, M);
    dfs(w * 2 + 2, M, R);
  }
  stk.pop_back();
}
int main() {
  std::cin >> N >> K;
  for (int i = 0; i < N; i++) {
    int V, W;
    std::cin >> V >> W;
    exh.emplace_back(V, W);
    begin.push_back(0);
    end.push_back(1e9);
  }
  int Q;
  std::cin >> Q;
  for (int i = 0; i < Q; i++) {
    int type;
    std::cin >> type;
    switch (type) {
      case 1: {
        int V, W;
        std::cin >> V >> W;
        exh.emplace_back(V, W);
        begin.push_back(i);
        end.push_back(1e9);
        break;
      }
      case 2: {
        int X;
        std::cin >> X;
        X--;
        end[X] = i;
        break;
      }
      case 3: {
        query(0, 0, Q, i);
        break;
      }
    }
  }
  for (int i = 0; i < exh.size(); i++) {
    insert(0, 0, Q, begin[i], end[i], exh[i]);
  }
  stk.emplace_back();
  dfs(0, 0, Q);
  return 0;
}