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>
const int oo = 0x3f3f3f3f;
template <typename T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return a > b ? a = b, true : false;
}
template <typename T>
T read(T &first) {
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (first = 0; isdigit(ch); ch = getchar()) first = 10 * first + ch - '0';
  return first *= f;
}
template <typename T>
void write(T first) {
  if (first < 0) {
    putchar('-');
    first = -first;
  }
  static char s[20];
  int top = 0;
  for (; first; first /= 10) s[++top] = first % 10 + '0';
  if (top == 0)
    putchar('0');
  else
    while (top) putchar(s[top--]);
}
const int MAXN = 1e4 + 5e3 + 5, MAXK = 1e3 + 5, MAXQ = 3e4 + 5;
const int _____P = 1e7 + 19, Q_____ = 1e9 + 7;
int N, K, Q;
std::pair<int, int> A[MAXN];
int QRY[MAXQ];
std::pair<int, int> range[MAXN];
int dp[MAXN][MAXK];
int ans[MAXQ][MAXK];
struct Edge {
  int v, next;
};
namespace sora_chan_saikou_desu {
const int SIZE_V = MAXQ << 2, SIZE_E = MAXN * 20;
int tote, head[SIZE_V];
Edge edge[SIZE_E];
int L, R;
int X;
void _insert(int u, int l, int r) {
  if (L <= l && r <= R) {
    edge[++tote] = (Edge){X, head[u]};
    head[u] = tote;
    return;
  }
  int mid = (l + r) >> 1;
  if (L <= mid) _insert((u << 1), l, mid);
  if (mid < R) _insert((u << 1 | 1), mid + 1, r);
}
inline void insert(int l, int r, int first) {
  L = l;
  R = r;
  X = first;
  _insert(1, 1, Q);
}
}  // namespace sora_chan_saikou_desu
void calc(int u, int cur, int l, int r) {
  using sora_chan_saikou_desu::edge;
  using sora_chan_saikou_desu::head;
  int *f = dp[cur];
  memcpy(f, dp[cur - 1], sizeof(dp[cur]));
  for (int i = head[u]; i; i = edge[i].next) {
    int v = A[edge[i].v].first, w = A[edge[i].v].second;
    for (int j = K; j >= w; --j) {
      chkmax(f[j], f[j - w] + v);
    }
  }
  if (l == r) {
    memcpy(ans[l], f, sizeof(ans[l]));
  } else {
    int mid = (l + r) >> 1;
    calc((u << 1), cur + 1, l, mid);
    calc((u << 1 | 1), cur + 1, mid + 1, r);
  }
}
void input() {
  read(N);
  read(K);
  int N0 = N;
  for (int i = 1; i <= N; ++i) {
    read(A[i].first);
    read(A[i].second);
  }
  read(Q);
  std::fill(range + 1, range + 1 + N, std::make_pair(1, N0 + Q));
  int Q0 = 0;
  for (int i = 1; i <= Q; ++i) {
    int type;
    read(type);
    if (type == 1) {
      ++N;
      read(A[N].first);
      read(A[N].second);
      range[N] = std::make_pair(N0 + i, N0 + Q);
    } else if (type == 2) {
      int p;
      read(p);
      range[p].second = N0 + i;
    } else {
      QRY[++Q0] = N0 + i;
    }
  }
  Q = Q0;
}
void solve() {
  std::sort(QRY + 1, QRY + Q + 1);
  for (int i = 1; i <= N; ++i) {
    int l = std::lower_bound(QRY + 1, QRY + Q + 1, range[i].first) - QRY;
    int r = std::upper_bound(QRY + 1, QRY + Q + 1, range[i].second) - QRY - 1;
    if (l <= r) sora_chan_saikou_desu::insert(l, r, i);
  }
  calc(1, 1, 1, Q);
  for (int i = 1; i <= Q; ++i) {
    int sum = 0;
    long long pm = 1;
    for (int j = 1; j <= K; ++j) {
      sum += ans[i][j] * pm % Q_____;
      if (sum >= Q_____) sum -= Q_____;
      pm = pm * _____P % Q_____;
    }
    write(sum);
    putchar('\n');
  }
}
int main() {
  input();
  solve();
  return 0;
}