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>
template <typename T>
inline T abs(T a) {
  return a < 0 ? -a : a;
}
template <typename T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
template <typename T>
inline T max(T a, T b) {
  return a > b ? a : b;
}
template <typename T>
inline void swap(T& a, T& b) {
  T t = a;
  a = b, b = t;
}
template <typename T>
inline bool umin(T& mn, T u) {
  return u < mn ? mn = u, true : false;
}
template <typename T>
inline bool umax(T& mx, T u) {
  return u > mx ? mx = u, true : false;
}
template <typename T>
inline void read(T& r) {
  char c;
  r = 0;
  bool flag = false;
  do c = getchar();
  while (c != '-' && (c < '0' || c > '9'));
  if (c == '-') flag = true, c = getchar();
  do r = r * 10 + c - '0', c = getchar();
  while (c >= '0' && c <= '9');
  if (flag) r *= -1;
}
template <typename T1, typename T2>
inline void read(T1& r, T2& s) {
  read(r), read(s);
}
template <typename T1, typename T2, typename T3>
inline void read(T1& r, T2& s, T3& t) {
  read(r), read(s), read(t);
}
template <typename T1, typename T2, typename T3, typename T4>
inline void read(T1& r, T2& s, T3& t, T4& u) {
  read(r), read(s), read(t), read(u);
}
template <typename T>
inline void write(T w, const char* c = "") {
  if (w == 0)
    putchar('0');
  else {
    if (w < 0) putchar('-'), w *= -1;
    static int s[21], top;
    top = 0;
    while (w) s[top++] = w % 10, w /= 10;
    while (top--) putchar(s[top] + '0');
  }
  for (int i = 0; c[i]; i++) putchar(c[i]);
}
int n, k, qn, en;
long long p[10000];
const long long P = 1e7 + 19, mod = 1e9 + 7;
void init() {
  for (int i = p[0] = 1; i < k; i++) p[i] = p[i - 1] * P % mod;
}
struct knapsack {
  int dp[10001];
  void add(std::pair<int, int> o) {
    int v = o.first, w = o.second;
    for (int i = k - w; i >= 0; i--) umax(dp[i + w], dp[i] + v);
  }
  int getans() {
    long long ans = 0;
    for (int i = 1; i <= k; i++) (ans += dp[i] * p[i - 1]) %= mod;
    return (int)ans;
  }
} st[160];
struct perfect_binary_tree {
  struct ac {
    std::pair<int, int> to;
    int nx;
  } e[4500000];
  int L, first[655360];
  perfect_binary_tree() : L(1) {}
  inline void addedge(int u, std::pair<int, int> v) {
    e[L].to = v, e[L].nx = first[u], first[u] = L++;
  }
  void add(int L, int R, std::pair<int, int> c, int now = 1, int l = 1,
           int r = qn) {
    if (L <= l && r <= R) {
      addedge(now, c);
      return;
    }
    int m = (l + r) >> 1;
    if (L <= m) add(L, R, c, now << 1, l, m);
    if (m < R) add(L, R, c, now << 1 | 1, m + 1, r);
  }
  void getans(int depth = 1, int now = 1, int l = 1, int r = qn) {
    st[depth] = st[depth - 1];
    for (int j = first[now]; j; j = e[j].nx) st[depth].add(e[j].to);
    if (l == r) {
      write(st[depth].getans(), "\n");
      return;
    }
    int m = (l + r) >> 1;
    getans(depth + 1, now << 1, l, m);
    getans(depth + 1, now << 1 | 1, m + 1, r);
  }
} T;
std::pair<int, int> exhibit[150000], show[150000];
void solve() {
  init();
  for (int i = 0; i < n; i++)
    if (show[i].first <= qn && show[i].second >= 1)
      T.add(show[i].first, show[i].second, exhibit[i]);
  T.getans();
}
void input() {
  qn = 0, n = 0;
  int m, o;
  read(m, k);
  while (m--)
    read(exhibit[n].first, exhibit[n].second),
        show[n++] = std::pair<int, int>(0, -1);
  read(m);
  while (m--) {
    read(o);
    switch (o) {
      case 1:
        read(exhibit[n].first, exhibit[n].second),
            show[n++] = std::pair<int, int>(qn + 1, -1);
        break;
      case 2:
        read(o), show[--o].second = qn;
        break;
      case 3:
        ++qn;
        break;
    }
  }
  for (int i = 0; i < n; i++)
    if (show[i].second == -1) show[i].second = qn;
}
int main() {
  input();
  solve();
  return 0;
}