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;
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << ", ";
  err(++it, args...);
}
template <typename T>
struct number_iterator : std::iterator<random_access_iterator_tag, T> {
  T v;
  number_iterator(T _v) : v(_v) {}
  operator T&() { return v; }
  T operator*() const { return v; }
};
template <typename T>
struct number_range {
  T b, e;
  number_range(T b, T e) : b(b), e(e) {}
  number_iterator<T> begin() { return b; }
  number_iterator<T> end() { return e; }
};
template <typename T>
number_range<T> range(T e) {
  return number_range<T>(1, e + 1);
}
template <typename T>
number_range<T> range(T b, T e) {
  return number_range<T>(b, e + 1);
}
int l[500001], r[500001], h, cnt, v[500001], w[500001];
int n, lb[500001], rb[500001], q;
int k;
long long res[500001];
vector<int> Q, a[500001][2];
void build(int u, int lb, int rb) {
  if (lb == rb) return;
  build(l[u] = ++h, lb, (lb + rb) / 2);
  build(r[u] = ++h, (lb + rb) / 2 + 1, rb);
}
void update(int u, int lb, int rb, int lq, int rq, int v1, int v2) {
  if (lb >= lq && rb <= rq) {
    a[u][0].push_back(v1);
    a[u][1].push_back(v2);
    return;
  }
  int mid = (lb + rb) / 2;
  if (mid >= lq) update(l[u], lb, mid, lq, rq, v1, v2);
  if (mid < rq) update(r[u], mid + 1, rb, lq, rq, v1, v2);
}
void solve(int u, int lb, int rb, vector<long long> dp) {
  for (int i = 0; i < a[u][0].size(); i++) {
    int v = a[u][0][i], w = a[u][1][i];
    for (int j = k; j >= w; j--) dp[j] = max(dp[j], dp[j - w] + v);
  }
  if (lb == rb) {
    res[lb] = 0;
    for (int j = k; j >= 1; j--) {
      res[lb] = (res[lb] * (long long)(10000000 + 19) + dp[j]) %
                (long long)(1000000000 + 7);
    }
  } else {
    solve(l[u], lb, (lb + rb) / 2, dp);
    solve(r[u], (lb + rb) / 2 + 1, rb, dp);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (auto i : range(n)) {
    scanf("%d%d", &v[i], &w[i]);
    lb[i] = 1;
  }
  cnt = n;
  scanf("%d", &q);
  for (auto i : range(n)) rb[i] = q;
  for (auto i : range(q)) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      cnt++;
      scanf("%d%d", &v[cnt], &w[cnt]);
      lb[cnt] = i;
      rb[cnt] = q;
    } else if (op == 2) {
      int t;
      scanf("%d", &t);
      rb[t] = i;
    } else {
      Q.push_back(i);
    }
  }
  build(h = 0, 1, q);
  for (auto i : range(cnt)) update(0, 1, q, lb[i], rb[i], v[i], w[i]);
  vector<long long> dp;
  for (auto i : range(0, k)) dp.push_back(0);
  solve(0, 1, q, dp);
  for (auto i : Q) printf("%I64d\n", res[i]);
  return 0;
}
