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;
template <class T>
void Read(T& x) {
  x = 0;
  int flag = 0, sgn = 1;
  char c;
  while (c = getchar()) {
    if (c == '-')
      sgn = -1;
    else if (c >= '0' && c <= '9')
      x *= 10, x += c - '0', flag = 1;
    else if (flag)
      break;
  }
  x *= sgn;
}
const int MAXN = 20000, MAXK = 1010;
const int p = 1e7 + 19, q = 1e9 + 7;
pair<int, int> vm[MAXN], zeit[MAXN];
int dp[30][MAXK], n, k, Q;
vector<int> query;
int tot = 0;
struct Segment_Tree {
  int first[40 * MAXN], next[40 * MAXN], e, to[40 * MAXN];
  Segment_Tree() {
    memset(first, -1, sizeof(first));
    e = 0;
  }
  void addEdge(int u, int v) {
    ++e;
    next[e] = first[u];
    first[u] = e;
    to[e] = v;
  }
  void add(int now, int l, int r, int ql, int qr, int v) {
    if (ql > qr) return;
    if (ql <= l && qr >= r) {
      addEdge(now, v);
      return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) add(now << 1, l, mid, ql, qr, v);
    if (qr > mid) add(now << 1 | 1, mid + 1, r, ql, qr, v);
  }
  void solve(int now, int l, int r, int dep) {
    memcpy(dp[dep], dp[dep - 1], sizeof(dp[dep - 1]));
    for (int i = first[now]; i != -1; i = next[i]) {
      int v = to[i];
      for (int j = k; j >= vm[v].second; j--)
        dp[dep][j] = max(dp[dep][j], dp[dep][j - vm[v].second] + vm[v].first);
    }
    if (l == r) {
      long long ans = 0, tmp = 1;
      for (int i = 1; i <= k; i++) {
        ans = (ans + dp[dep][i] * tmp % q) % q;
        tmp = tmp * p % q;
      }
      printf("%d\n", ans);
      return;
    }
    int mid = (l + r) >> 1;
    solve(now << 1, l, mid, dep + 1);
    solve(now << 1 | 1, mid + 1, r, dep + 1);
  }
} T;
int main() {
  Read(n), Read(k);
  for (int i = 1; i <= n; i++) {
    ++tot;
    int v, w;
    Read(v), Read(w);
    vm[tot].first = v, vm[tot].second = w;
    zeit[tot].first = 1;
    zeit[tot].second = -1;
  }
  Read(Q);
  int lover = 0;
  for (int i = 1; i <= Q; i++) {
    int op;
    Read(op);
    if (op == 1) {
      int v, w;
      Read(v), Read(w);
      vm[++tot] = make_pair(v, w);
      zeit[tot].first = lover + 1;
      zeit[tot].second = -1;
    } else if (op == 2) {
      int x;
      Read(x);
      zeit[x].second = lover;
    } else
      lover++;
  }
  for (int i = 1; i <= tot; i++) {
    if (zeit[i].second == -1) zeit[i].second = lover;
    T.add(1, 1, lover, zeit[i].first, zeit[i].second, i);
  }
  T.solve(1, 1, lover, 1);
}
