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 Mo = 1000000007LL;
int f[21][1010];
bitset<15010> sub[101000], now;
int v[30100], w[30100], lk[101000], rk[101000];
int n, m, K, top, N, T, root;
long long fac[1010];
int build(int L, int R) {
  if (L == R) return L;
  int u = ++top;
  lk[u] = build(L, ((L + R) >> 1));
  rk[u] = build((((L + R) >> 1) + 1), R);
  sub[u] = sub[lk[u]] & sub[rk[u]];
  return u;
}
void solve(int dep, int fa, int u) {
  for (int i = 0; i <= K; i++) f[dep][i] = f[dep - 1][i];
  for (int i = 1; i <= n; i++) {
    if (!(sub[u][i] && !sub[fa][i])) continue;
    for (int j = K - w[i]; j >= 0; j--) {
      f[dep][j + w[i]] = max(f[dep][j + w[i]], f[dep][j] + v[i]);
    }
  }
  if (u <= N) {
    long long ans = 0;
    for (int i = 1; i <= K; i++) {
      ans += (long long)f[dep][i] * fac[i - 1];
      ans %= Mo;
    }
    printf("%I64d\n", ans);
    return;
  }
  solve(dep + 1, u, lk[u]);
  solve(dep + 1, u, rk[u]);
  return;
}
int main() {
  scanf("%d %d", &n, &K);
  fac[0] = 1LL;
  for (int i = 1; i <= K; i++) fac[i] = (fac[i - 1] * 10000019LL) % Mo;
  for (int i = 1; i <= n; i++) scanf("%d %d", &v[i], &w[i]);
  for (int i = 1; i <= n; i++) now[i] = 1;
  for (scanf("%d", &T); T; T--) {
    int k;
    scanf("%d", &k);
    if (k == 1) {
      n++;
      scanf("%d %d", &v[n], &w[n]);
      now[n] = 1;
    } else if (k == 2) {
      scanf("%d", &k);
      now[k] = 0;
    } else {
      sub[++N] = now;
    }
  }
  top = N;
  root = build(1, N);
  solve(1, 0, root);
  return 0;
}
