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 p = 1e7 + 19;
const int mod = 1e9 + 7;
vector<int> G[MAXN << 2];
int dp[30][1005];
int st[MAXN];
int ed[MAXN];
int op[MAXN];
int w[MAXN];
int v[MAXN];
int n, V, q;
int cnt;
void upd(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void update(int L, int R, int idx, int o, int l, int r) {
  if (L <= l && r <= R) {
    G[o].push_back(idx);
    return;
  }
  int m = ((l + r) >> 1);
  if (L <= m) update(L, R, idx, (o << 1), l, m);
  if (m < R) update(L, R, idx, (o << 1 | 1), m + 1, r);
}
void down(int d, int o, int l, int r) {
  for (int i = 1; i <= V; ++i) {
    dp[d][i] = dp[d - 1][i];
  }
  for (int i = 0; i < G[o].size(); ++i) {
    int x = G[o][i], cost = v[x], val = w[x];
    for (int j = V; j >= cost; --j) {
      dp[d][j] = max(dp[d][j], dp[d][j - cost] + val);
    }
  }
  if (l == r) {
    int ans = 0, t = 1;
    if (op[l] == 3) {
      for (int i = 1; i <= V; ++i) {
        upd(ans, 1LL * t * dp[d][i] % mod);
        t = 1LL * t * p % mod;
      }
      printf("%d\n", ans);
    }
    return;
  }
  int m = ((l + r) >> 1);
  down(d + 1, (o << 1), l, m);
  down(d + 1, (o << 1 | 1), m + 1, r);
}
void solve() {
  int x, y;
  cnt = n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &w[i], &v[i]);
  }
  scanf("%d", &q);
  for (int i = 1; i <= n; ++i) {
    st[i] = 1;
    ed[i] = q;
  }
  for (int i = 1; i <= q; ++i) {
    scanf("%d", &op[i]);
    if (op[i] == 1) {
      scanf("%d%d", &x, &y);
      ++cnt;
      w[cnt] = x;
      v[cnt] = y;
      st[cnt] = i;
      ed[cnt] = q;
    } else if (op[i] == 2) {
      scanf("%d", &x);
      ed[x] = i;
    }
  }
  for (int i = 0; i < q << 2; ++i) {
    G[i].clear();
  }
  for (int i = 1; i <= cnt; ++i) {
    update(st[i], ed[i], i, 1, 1, q);
  }
  memset(dp, 0, sizeof dp);
  down(1, 1, 1, q);
}
int main() {
  while (~scanf("%d%d", &n, &V)) solve();
  return 0;
}
