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>
const int maxn = 50050, logN = 18, maxk = 1005;
const int Mod = 1e9 + 7, base = 1e7 + 19;
int N, K, Q;
bool ext[maxn << 2];
int head[maxn << 2], size;
std::pair<int, int> point[maxn * logN], item[maxn];
int bag[logN][maxk];
int type[maxn], st[maxn], ed[maxn], tot;
int ans[maxn], L;
int newitem(int id, int x) {
  point[++size] = std::make_pair(x, head[id]), head[id] = size;
}
void build(int l, int r, int p) {
  if (l == r)
    ext[p] = type[l] == 3;
  else {
    int mid = (l + r) >> 1;
    build(l, mid, p << 1);
    build(mid + 1, r, p << 1 | 1);
    ext[p] = ext[p << 1] || ext[p << 1 | 1];
  }
}
void insert(int l, int r, int it, int ll, int rr, int p) {
  if (l == ll && r == rr)
    newitem(p, it);
  else {
    int mid = (ll + rr) >> 1;
    if (l > mid)
      insert(l, r, it, mid + 1, rr, p << 1 | 1);
    else if (r <= mid)
      insert(l, r, it, ll, mid, p << 1);
    else {
      insert(l, mid, it, ll, mid, p << 1);
      insert(mid + 1, r, it, mid + 1, rr, p << 1 | 1);
    }
  }
}
int count(int *A) {
  long long p = 1, ret = 0;
  for (int i = 1; i <= K; i++) {
    ret += p * A[i] % Mod, (p *= base) %= Mod;
  }
  return (int)(ret % Mod);
}
void add(int s, int *A) {
  int v = item[s].first, w = item[s].second;
  for (int i = K; i >= w; i--) A[i] = std::max(A[i - w] + v, A[i]);
}
void solve(int l, int r, int p, int dep) {
  if (ext[p] == 0) return;
  int *F = bag[dep], *G = bag[dep - 1];
  for (int i = 0; i <= K; i++) F[i] = G[i];
  for (int i = head[p]; i; i = point[i].second) add(point[i].first, F);
  if (l == r)
    ans[++L] = count(F);
  else {
    int mid = (l + r) >> 1;
    solve(l, mid, p << 1, dep + 1);
    solve(mid + 1, r, p << 1 | 1, dep + 1);
  }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1, v, w; i <= N; i++) {
    scanf("%d%d", &v, &w);
    item[i] = std::make_pair(v, w);
  }
  tot = N;
  scanf("%d", &Q);
  for (int i = 1; i <= N; i++) st[i] = 1, ed[i] = Q;
  for (int i = 1, x, y; i <= Q; i++) {
    scanf("%d", &type[i]);
    if (type[i] == 1) {
      scanf("%d%d", &x, &y);
      item[++tot] = std::make_pair(x, y);
      st[tot] = i, ed[tot] = Q;
    } else if (type[i] == 2) {
      scanf("%d", &x);
      ed[x] = i;
    }
  }
  build(1, Q, 1);
  for (int i = 1; i <= tot; i++) {
    if (i <= N)
      insert(st[i], ed[i], i, 1, Q, 1);
    else
      insert(st[i], ed[i], i, 1, Q, 1);
  }
  solve(1, Q, 1, 1);
  for (int i = 1; i <= L; i++) printf("%d\n", ans[i]);
  return 0;
}
