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;
const int maxn = 4e4 + 10;
const int maxm = 1010;
const int base = 31337;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int logo = 16;
const int off = 1 << logo;
const int treesiz = off << 1;
const int rn = 1e7 + 19;
int n, k, q;
int tour[treesiz];
vector<int> v[treesiz];
int val[maxn], w[maxn];
bool bio[maxn];
int f[maxn];
int sol[maxn];
int mul(int a, int b) {
  long long out = (long long)a * b;
  return out % mod;
}
void update(int a, int b, int l, int r, int node, int x) {
  if (l > b || r < a) return;
  if (l >= a && r <= b) {
    v[node].push_back(x);
    return;
  }
  int mid = (l + r) / 2;
  update(a, b, l, mid, node * 2, x);
  update(a, b, mid + 1, r, node * 2 + 1, x);
}
void build(int node, vector<int> dp) {
  if (node >= treesiz) return;
  for (int i = 0; i < v[node].size(); i++) {
    int va = val[v[node][i]];
    int we = w[v[node][i]];
    for (int j = k; j >= we; j--) {
      dp[j] = max(dp[j], dp[j - 1]);
      dp[j] = max(dp[j], dp[j - we] + va);
    }
  }
  if (tour[node] != -1) {
    int s = 0;
    int pot = 1;
    for (int i = 1; i <= k; i++) {
      s += mul(dp[i], pot);
      pot = mul(pot, rn);
      s %= mod;
    }
    sol[tour[node]] = s;
  }
  build(node * 2, dp);
  build(node * 2 + 1, dp);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", val + i, w + i);
    f[i] = i;
  }
  int ptr = n, qq = 0;
  memset(bio, false, sizeof bio);
  memset(tour, -1, sizeof tour);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      val[i + n] = x, w[i + n] = y;
      f[ptr++] = i + n;
    } else if (type == 2) {
      int x;
      scanf("%d", &x);
      x--;
      x = f[x];
      bio[x] = true;
      update(x, i + n, 0, off - 1, 1, x);
    } else {
      tour[i + n + off] = qq++;
    }
  }
  for (int i = 0; i < n + q; i++) {
    if (!bio[i] && w[i] > 0) {
      update(i, n + q, 0, off - 1, 1, i);
    }
  }
  vector<int> dp;
  for (int i = 0; i <= k; i++) dp.push_back(0);
  build(1, dp);
  for (int i = 0; i < qq; i++) printf("%d\n", sol[i]);
  return 0;
}
