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;
struct pp {
  int lc, rc;
} p[60010];
int root, pl;
int f[30][1010], K;
int m1 = 10000019, mod = 1000000007;
int L[15010], R[15010], v[15010], w[15010];
vector<int> z[60010];
int ans[30010];
void build(int &x, int l, int r, int ll, int rr, int k) {
  if (!x) x = ++pl;
  if (l == ll && r == rr) {
    z[x].push_back(k);
    return;
  }
  int mid = (l + r) >> 1;
  if (rr <= mid)
    build(p[x].lc, l, mid, ll, rr, k);
  else if (ll > mid)
    build(p[x].rc, mid + 1, r, ll, rr, k);
  else {
    build(p[x].lc, l, mid, ll, mid, k);
    build(p[x].rc, mid + 1, r, mid + 1, rr, k);
  }
}
void dfs(int x, int l, int r, int d) {
  int i, j;
  for (i = 0; i <= K; i++) f[d][i] = f[d - 1][i];
  for (i = 0; i < z[x].size(); i++) {
    int W = w[z[x][i]], V = v[z[x][i]];
    for (j = K; j >= W; j--) f[d][j] = max(f[d][j], f[d][j - W] + V);
  }
  if (l == r)
    for (i = K; i >= 1; i--) ans[l] = ((long long)ans[l] * m1 + f[d][i]) % mod;
  else {
    int mid = (l + r) >> 1;
    dfs(p[x].lc, l, mid, d + 1);
    dfs(p[x].rc, mid + 1, r, d + 1);
  }
}
int main() {
  int n, ql = 0, m, t, x, i, j;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &v[i], &w[i]);
    L[i] = 1;
    R[i] = -1;
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &t);
    if (t == 1) {
      L[++n] = ql + 1;
      R[n] = -1;
      scanf("%d%d", &v[n], &w[n]);
    } else if (t == 2) {
      scanf("%d", &x);
      R[x] = ql;
    } else
      ql++;
  }
  for (i = 1; i <= n; i++) {
    if (R[i] == -1) R[i] = ql;
    if (L[i] <= R[i]) build(root, 1, ql, L[i], R[i], i);
  }
  dfs(root, 1, ql, 1);
  for (i = 1; i <= ql; i++) printf("%d\n", ans[i]);
  return 0;
}
