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 N = 40005;
const int M = 1005;
const int Q = 1000000007;
const int P = 10000019;
int n, m, q, tot;
int v[N], w[N], Ask[N], f[N], id[N];
int g[M], dp[N * 4];
vector<int> T[N * 4];
void Build(int s, int x, int y, int l, int r, int k) {
  if (l <= x && y <= r) {
    T[s].push_back(k);
    return;
  }
  int Mid = (x + y) / 2;
  int t = s + s;
  if (l <= Mid) Build(t, x, Mid, l, r, k);
  if (Mid < r) Build(t + 1, Mid + 1, y, l, r, k);
}
void Make(int s, int x, int y) {
  if (x == y) {
    dp[s] = Ask[x];
    return;
  }
  int Mid = (x + y) / 2;
  int t = s + s;
  Make(t, x, Mid);
  Make(t + 1, Mid + 1, y);
  dp[s] = dp[t] | dp[t + 1];
}
void Add(int dp[M], int y, int x) {
  for (int i = m; i >= x; i--) dp[i] = max(dp[i], dp[i - x] + y);
}
void print() {
  int Ans = 0;
  for (int i = m; i; i--) Ans = (1LL * P * Ans + g[i]) % Q;
  printf("%d\n", Ans);
}
void Work(int s, int x, int y) {
  if (!dp[s]) return;
  int Mid = (x + y) / 2;
  int tmp[N];
  for (int i = 0; i <= m; i++) tmp[i] = g[i];
  for (int j = 0; j < T[s].size(); j++) Add(g, v[T[s][j]], w[T[s][j]]);
  if (x == y) {
    print();
    for (int i = 0; i <= m; i++) g[i] = tmp[i];
    return;
  }
  int t = s + s;
  Work(t, x, Mid);
  Work(t + 1, Mid + 1, y);
  for (int i = 0; i <= m; i++) g[i] = tmp[i];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &v[i], &w[i]);
    f[i] = 1;
    id[i] = i;
  }
  scanf("%d", &q);
  tot = n;
  for (int i = 1; i <= q; i++) {
    int x;
    scanf("%d", &x);
    if (x == 1) {
      tot++;
      scanf("%d%d", &v[tot], &w[tot]);
      f[tot] = 1;
      id[tot] = n + i;
    } else if (x == 2) {
      int y;
      scanf("%d", &y);
      Build(1, 1, n + q, id[y], n + i, y);
      f[y] = 0;
    } else if (x == 3) {
      Ask[n + i] = 1;
    }
  }
  for (int i = 1; i <= n; i++)
    if (f[i]) Add(g, v[i], w[i]);
  for (int i = n + 1; i <= tot; i++)
    if (f[i]) Build(1, 1, n + q, id[i], n + q, i);
  Make(1, 1, n + q);
  Work(1, 1, n + q);
  return 0;
}
