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>
using namespace std;
template <class T>
inline void Read(T &x) {
  int f = 1;
  char t = getchar();
  while (t < '0' || t > '9') {
    if (t == '-') f = -1;
    t = getchar();
  }
  x = 0;
  while (t >= '0' && t <= '9') {
    x = x * 10 + t - '0';
    t = getchar();
  }
  x *= f;
}
const int logn = 25;
const int maxv = 1005;
const int maxn = 20005;
const int maxq = 30005;
const int seed = 1e7 + 19;
const int modulo = 1e9 + 7;
int n, k, q, size;
int w[maxn], v[maxn];
int intime[maxn], outtime[maxn];
vector<int> query;
vector<int> T[maxq << 2];
int f[logn][maxn], ans[maxq], p[maxv];
int fpm(int x, int k) {
  int re = 1;
  while (k > 0) {
    if (k & 1) re = 1ll * re * x % modulo;
    x = 1ll * x * x % modulo;
    k >>= 1;
  }
  return re;
}
void input() {
  Read(n), Read(k);
  for (register int i = 1; i <= n; i++) {
    Read(w[i]), Read(v[i]);
    intime[i] = 1;
  }
  Read(q);
  size = sizeof(int) * (k + 1);
}
void insert(int cn, int l, int r, int l0, int r0, int id) {
  if (l == l0 && r == r0) {
    T[cn].push_back(id);
    return;
  }
  int mid = (l + r) >> 1;
  if (r0 <= mid)
    insert(cn << 1, l, mid, l0, r0, id);
  else if (l0 > mid)
    insert(cn << 1 | 1, mid + 1, r, l0, r0, id);
  else
    insert(cn << 1, l, mid, l0, mid, id),
        insert(cn << 1 | 1, mid + 1, r, mid + 1, r0, id);
}
void dfs(int cn, int l, int r, int dep) {
  if (dep == 1)
    memset(f[dep], 0, size);
  else
    memcpy(f[dep], f[dep - 1], size);
  int now, x, y;
  for (register int i = 0; i < T[cn].size(); i++) {
    now = T[cn][i], x = w[now], y = v[now];
    for (register int j = k; j >= y; j--) {
      f[dep][j] = max(f[dep][j], f[dep][j - y] + x);
    }
  }
  if (l == r) {
    int cnt = 0;
    for (register int i = 1; i <= k; i++)
      cnt = (cnt + 1ll * p[i] * f[dep][i] % modulo) % modulo;
    ans[l] = cnt;
    return;
  }
  int mid = (l + r) >> 1;
  dfs(cn << 1, l, mid, dep + 1);
  dfs(cn << 1 | 1, mid + 1, r, dep + 1);
}
void prepare() {
  int op, x, y;
  q++;
  for (register int i = 2; i <= q; i++) {
    Read(op);
    if (op == 1) {
      Read(x), Read(y);
      w[++n] = x, v[n] = y;
      intime[n] = i;
    } else if (op == 2) {
      Read(x);
      outtime[x] = i;
    } else {
      query.push_back(i);
    }
  }
  for (register int i = 1; i <= n; i++)
    if (!outtime[i]) outtime[i] = q;
  for (register int i = 1; i <= n; i++) {
    insert(1, 1, q, intime[i], outtime[i], i);
  }
  for (register int i = 1; i <= k; i++) p[i] = fpm(seed, i - 1);
  dfs(1, 1, q, 1);
}
void solve() {
  for (register int i = 0; i < query.size(); i++) {
    printf("%d\n", ans[query[i]]);
  }
}
int main() {
  input();
  prepare();
  solve();
  return 0;
}