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>
template <typename T>
bool Chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool Chkmin(T &x, T y) {
  return y < x ? x = y, true : false;
}
const int MAXN = 3e4 + 1e3, MAXNODE = MAXN * 20, MAXEDGE = MAXN * 20,
          MAXK = 1e3 + 1e2, PP = 1e7 + 19, MOD = 1e9 + 7, oo = 0x3f3f3f3f;
int n, k, q, totq;
int be[MAXNODE], ed[MAXNODE];
int w[MAXN];
int val[MAXN];
int head[MAXNODE], next[MAXEDGE];
int to[MAXEDGE];
int ans[MAXN];
int top;
int dp[20][MAXNODE];
inline void Mod(int &x) { x = (x % MOD + MOD) % MOD; }
inline void AddEdge(int u, int v) {
  static int tote;
  ++tote;
  to[tote] = v;
  next[tote] = head[u];
  head[u] = tote;
}
namespace SMT {
void Exec(int o = 1, int l = 1, int r = totq) {
  ++top;
  int *f = dp[top];
  for (int i = 0, _ = k; i <= _; i++) f[i] = dp[top - 1][i];
  for (int i = head[o]; i; i = next[i]) {
    int v = to[i];
    for (int j = k, _ = w[v]; j >= _; j--) Chkmax(f[j], f[j - w[v]] + val[v]);
  }
  int mid = (l + r) / 2;
  if (l != r) {
    Exec((o * 2), l, mid);
    Exec((o * 2 + 1), mid + 1, r);
  } else {
    int tmp = 1;
    for (int i = 1, _ = k; i <= _; i++) {
      Mod(ans[mid] += (long long)f[i] * tmp % MOD);
      tmp = (long long)tmp * PP % MOD;
    }
  }
  --top;
}
void Add(int o, int l, int r, int L, int R, int X) {
  int mid = (l + r) / 2;
  if (L <= l && r <= R)
    AddEdge(o, X);
  else {
    if (L <= mid) Add((o * 2), l, mid, L, R, X);
    if (mid < R) Add((o * 2 + 1), mid + 1, r, L, R, X);
  }
}
inline void Add(int l, int r, int x) {
  if (l <= r) Add(1, 1, totq, l, r, x);
}
}  // namespace SMT
void Solve() {
  for (int i = 1, _ = n; i <= _; i++) {
    SMT::Add(be[i], ed[i], i);
  }
  SMT::Exec();
  for (int i = 1, _ = totq; i <= _; i++) printf("%d\n", ans[i]);
}
void Input() {
  scanf("%d%d", &n, &k);
  for (int i = 1, _ = n; i <= _; i++) {
    scanf("%d%d", &val[i], &w[i]);
    be[i] = 1, ed[i] = oo;
  }
  scanf("%d", &q);
  for (int i = 1, _ = q; i <= _; i++) {
    int type, x;
    scanf("%d", &type);
    if (type == 1) {
      ++n;
      scanf("%d%d", &val[n], &w[n]);
      be[n] = totq + 1, ed[n] = oo;
    } else if (type == 2) {
      scanf("%d", &x);
      ed[x] = totq;
    } else
      ++totq;
  }
}
int main() {
  if (fopen("E.in", "r") != NULL) {
    freopen("E.in", "r", stdin);
    freopen("E.out", "w", stdout);
  }
  Input();
  Solve();
  return 0;
}