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, P = 1e7 + 19, Q = 1e9 + 7;
int seg[maxn << 2], *dp[maxn << 2], obj[maxn * 17], nxt[maxn * 17], cnt;
void ins(int u, int v) {
  obj[++cnt] = v;
  nxt[cnt] = seg[u];
  seg[u] = cnt;
}
int i, j, tot, n, m, lx[maxn], rx[maxn], wt[maxn], va[maxn];
int qry[maxn], qcnt;
void upd(int x, int l, int r, int L, int R, int OBJ) {
  if (L <= l && r <= R) {
    ins(x, OBJ);
    return;
  }
  int mid = (l + r) >> 1;
  if (L <= mid) upd(x << 1, l, mid, L, R, OBJ);
  if (R > mid) upd(x << 1 | 1, mid + 1, r, L, R, OBJ);
}
void Max(int &x, int y) {
  if (x < y) x = y;
}
int cal(int x, int l, int r, int tar) {
  if (seg[x] != -1) {
    int *pd = dp[x] = new int[m + 1], i, j, t;
    memset(pd, 0, (m + 1) << 2);
    if (x != 1) memcpy(pd, dp[x >> 1], (m + 1) << 2);
    for (i = seg[x]; t = obj[i], i; i = nxt[i]) {
      for (j = m; j >= wt[t]; j--) Max(pd[j], pd[j - wt[t]] + va[t]);
    }
    seg[x] = -1;
  }
  if (l == r) {
    int ans = 0, i;
    for (i = m; i >= 1; i--) ans = ((long long)ans * P + dp[x][i]) % Q;
    return ans;
  }
  int mid = (l + r) >> 1;
  if (tar <= mid) return cal(x << 1, l, mid, tar);
  return cal(x << 1 | 1, mid + 1, r, tar);
}
int main() {
  scanf("%d%d", &tot, &m);
  for (i = 1; i <= tot; i++) scanf("%d%d", va + i, wt + i);
  scanf("%d", &n);
  for (i = 1; i <= tot; i++) lx[i] = 1, rx[i] = n;
  int opt, v, w;
  for (i = 1; i <= n; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d", &v, &w);
      ++tot;
      lx[tot] = i;
      rx[tot] = n;
      va[tot] = v;
      wt[tot] = w;
    } else if (opt == 2) {
      scanf("%d", &v);
      rx[v] = i;
    } else
      qry[++qcnt] = i;
  }
  for (i = 1; i <= tot; i++) upd(1, 1, n, lx[i], rx[i], i);
  for (i = 1; i <= qcnt; i++) printf("%d\n", cal(1, 1, n, qry[i]));
  return 0;
}
