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 = 30005;
const int maxm = 1005;
int P = 1e7 + 19;
int Q = 1e9 + 7;
int n, m, K;
struct Thing {
  int l, r;
  int val, wei;
  Thing() { l = r = 0; }
} A[maxn];
int ask[maxn] = {0}, at = 0;
long long f[maxm][maxm] = {0};
struct edge {
  int to, next;
  edge() {}
  edge(int _t, int _n) : to(_t), next(_n) {}
} e[maxn * 17];
int head[maxn * 17] = {0}, et = -1;
void Init() {
  register int kind;
  register int x, v, w;
  scanf("%d%d", &n, &K);
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &A[i].val, &A[i].wei);
    A[i].l = 0;
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &kind);
    if (kind == 1) {
      scanf("%d%d", &v, &w);
      ++n;
      A[n].l = i;
      A[n].val = v;
      A[n].wei = w;
    }
    if (kind == 2) {
      scanf("%d", &x);
      A[x].r = i;
    }
    if (kind == 3) {
      ask[++at] = i;
    }
  }
}
void Insert(int L, int R, int l, int r, int tip, int p) {
  int mid = (L + R) / 2;
  if (l <= L && r >= R) {
    e[++et] = edge(tip, head[p]), head[p] = et;
    return;
  }
  if (l <= mid) Insert(L, mid, l, r, tip, p << 1);
  if (r > mid) Insert(mid + 1, R, l, r, tip, p << 1 | 1);
}
void Prepare() {
  register int L, R;
  for (int i = 1; i <= n; ++i) {
    if (A[i].r == 0) {
      A[i].r = m + 1;
    }
    L = lower_bound(ask + 1, ask + 1 + at, A[i].l) - ask;
    R = lower_bound(ask + 1, ask + 1 + at, A[i].r) - ask - 1;
    if (R < L) continue;
    Insert(1, at, L, R, i, 1);
  }
}
void Dfs(int L, int R, int p, int deep) {
  int mid = (L + R) / 2;
  memcpy(f[deep], f[deep - 1], sizeof(f[deep]));
  for (int i = head[p]; i != -1; i = e[i].next) {
    int cost = A[e[i].to].wei;
    int val = A[e[i].to].val;
    for (int j = K - cost; j >= 0; --j) {
      f[deep][j + cost] = max(f[deep][j + cost], f[deep][j] + val);
    }
  }
  if (L == R) {
    long long S = 0;
    for (int i = K; i >= 1; --i) {
      S = S * P;
      S = (S + f[deep][i]) % Q;
    }
    printf("%I64d\n", S);
    return;
  }
  Dfs(L, mid, p << 1, deep + 1);
  Dfs(mid + 1, R, p << 1 | 1, deep + 1);
}
void Solve() { Dfs(1, at, 1, 1); }
int main() {
  Init();
  Prepare();
  Solve();
  return 0;
}
