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 long long int MOD = 1000000007LL, BASE = 10000019LL;
struct sgtNode {
  long long int dp[1003];
  sgtNode() { memset(dp, 0, sizeof(dp)); }
};
typedef sgtNode *tnode;
tnode root[65537];
vector<int> tag[65537];
pair<int, int> exhibit[15005];
int n, m, ql, qr, qv, st[15005], ed[15005], curTime, Q, q[30005];
long long int *temp;
void sgtMark(int x, int l, int r) {
  if (l >= ql && r <= qr) {
    tag[x].push_back(qv);
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) sgtMark(x << 1, l, mid);
  if (qr > mid) sgtMark(x << 1 | 1, mid + 1, r);
}
long long int *sgtQuery(int x, int l, int r, int p) {
  if (!root[x]) {
    root[x] = new sgtNode;
    temp = root[x]->dp;
    if (x != 1) memcpy(temp, root[x >> 1]->dp, sizeof(sgtNode));
    while (!tag[x].empty()) {
      for (int i = m; i >= exhibit[tag[x].back()].first; --i)
        temp[i] = max(temp[i], temp[i - exhibit[tag[x].back()].first] +
                                   exhibit[tag[x].back()].second);
      tag[x].pop_back();
    }
  }
  if (l == r) return root[x]->dp;
  int mid = (l + r) >> 1;
  if (p <= mid) return sgtQuery(x << 1, l, mid, p);
  return sgtQuery(x << 1 | 1, mid + 1, r, p);
}
int main() {
  int i, j, op, t;
  long long int ans, base, *res;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; ++i) {
    scanf("%d%d", &exhibit[i].second, &exhibit[i].first);
    st[i] = i;
  }
  curTime = n;
  scanf("%d", &Q);
  for (i = 1; i <= Q; ++i) {
    scanf("%d", &op);
    if (op == 1) {
      st[++n] = ++curTime;
      scanf("%d%d", &exhibit[n].second, &exhibit[n].first);
    } else if (op == 2) {
      scanf("%d", &t);
      ed[t] = ++curTime;
    } else
      q[++q[0]] = curTime;
  }
  for (i = 1; i <= n; ++i) {
    if (!ed[i]) ed[i] = curTime + 1;
    ql = st[i], qr = ed[i] - 1, qv = i;
    sgtMark(1, 1, curTime);
  }
  for (i = 1; i <= q[0]; ++i) {
    res = sgtQuery(1, 1, curTime, q[i]);
    for (j = 1, base = 1LL, ans = 0LL; j <= m; ++j) {
      ans = (ans + ((res[j]) % MOD * base)) % MOD;
      base = (base * BASE) % MOD;
    }
    printf("%d\n", int(ans));
  }
  return 0;
}
