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;
const int maxn = 30000 + 10;
const int maxm = 600000 + 10;
const int P = 1e7 + 19;
const int Q = 1e9 + 7;
struct Edge {
  int next, v;
  Edge() {}
  Edge(int next, int v) : next(next), v(v) {}
} e[maxm];
struct Exhibit {
  int l, r, v, w;
  Exhibit() {
    l = 1;
    r = 100000;
  }
} a[maxn];
int n, m, q, K;
int head[maxm], cnt;
int dp[20][maxn];
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline void AddEdge(int u, int v) {
  e[++cnt] = Edge(head[u], v);
  head[u] = cnt;
}
void Insert(int o, int l, int r, int ql, int qr, int p) {
  if (ql <= l && r <= qr) {
    AddEdge(o, p);
    return;
  }
  int mid = (l + r) >> 1, lson = o << 1, rson = lson | 1;
  if (qr <= mid)
    Insert(lson, l, mid, ql, qr, p);
  else if (ql > mid)
    Insert(rson, mid + 1, r, ql, qr, p);
  else {
    Insert(lson, l, mid, ql, qr, p);
    Insert(rson, mid + 1, r, ql, qr, p);
  }
}
void solve(int o, int l, int r, int dep) {
  for (int i = 0; i <= K; i++) dp[dep][i] = dp[dep - 1][i];
  for (int i = head[o], val, w; i; i = e[i].next) {
    int v = e[i].v;
    val = a[v].v, w = a[v].w;
    for (int k = K - w; k >= 0; k--)
      dp[dep][k + w] = max(dp[dep][k + w], dp[dep][k] + val);
  }
  if (l == r) {
    long long res = 0;
    for (int i = K; i >= 1; i--) {
      res = res * P;
      res = (res + dp[dep][i]) % Q;
    }
    printf("%I64d\n", res);
    return;
  }
  int mid = (l + r) >> 1, lson = o << 1, rson = lson | 1;
  solve(lson, l, mid, dep + 1);
  solve(rson, mid + 1, r, dep + 1);
}
int main() {
  if (fopen("in.in", "r") != NULL) {
    freopen("in.in", "r", stdin);
    freopen("out.out", "w", stdout);
  }
  n = read(), K = read();
  for (int i = 1; i <= n; i++) a[i].v = read(), a[i].w = read();
  m = read();
  for (int i = 1, op; i <= m; i++) {
    op = read();
    if (op == 1) {
      n++;
      a[n].v = read(), a[n].w = read();
      a[n].l = q + 1;
    } else if (op == 2) {
      op = read();
      a[op].r = q;
    } else
      q++;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i].r == 100000) a[i].r = q;
    if (a[i].l <= a[i].r) Insert(1, 1, q, a[i].l, a[i].r, i);
  }
  solve(1, 1, q, 1);
  return 0;
}