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;
long long int start[100001], finish[100001], val[100001], wt[100001],
    store[100001], dp[100001], k, n, p = 10000019, prefix[100001], mark[100001],
                                     res[100001];
vector<pair<long long int, long long int> > tree[4 * 100001];
inline long long int add1(long long int x, long long int y) {
  return (x + y) % 1000000007;
}
inline long long int mul1(long long int x, long long int y) {
  return (x * y) % 1000000007;
}
void add(long long int node, long long int a, long long int b, long long int l,
         long long int r, long long int ind) {
  if (a > b || a > r || b < l) return;
  if (a >= l && b <= r) {
    tree[node].push_back({val[ind], wt[ind]});
    return;
  }
  long long int mid = (a + b) / 2;
  add(2 * node, a, mid, l, r, ind);
  add(2 * node + 1, mid + 1, b, l, r, ind);
}
void solve(long long int node, long long int a, long long int b) {
  if (a > b) return;
  long long int store[1001] = {0};
  for (long long int i = 0; i <= k; i++) store[i] = dp[i];
  for (auto it : tree[node]) {
    long long int value = it.first;
    long long int weight = it.second;
    for (long long int i = k; i >= 0; i--) {
      if (i + weight <= k) dp[i + weight] = max(dp[i + weight], dp[i] + value);
    }
  }
  prefix[0] = 0;
  for (long long int i = 1; i <= k; i++) prefix[i] = max(prefix[i - 1], dp[i]);
  if (a == b) {
    if (!mark[a]) {
      for (long long int i = 0; i <= k; i++) dp[i] = store[i];
      return;
    }
    long long int ans = 0;
    long long int temp = 1;
    for (long long int i = 1; i <= k; i++) {
      ans = add1(ans, mul1(prefix[i], temp));
      temp = mul1(temp, p);
    }
    res[a] = ans;
    for (long long int i = 0; i <= k; i++) dp[i] = store[i];
    return;
  }
  long long int mid = (a + b) / 2;
  solve(2 * node, a, mid);
  solve(2 * node + 1, mid + 1, b);
  for (long long int i = 0; i <= k; i++) dp[i] = store[i];
  return;
}
int main() {
  ios::sync_with_stdio(false);
  ;
  cin.tie(0);
  ;
  long long int ctr, q, type, x, y, i;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> val[i] >> wt[i];
    start[i] = 1;
  }
  ctr = n;
  cin >> q;
  for (i = 1; i <= q; i++) {
    cin >> type;
    if (type == 1) {
      ctr++;
      cin >> val[ctr] >> wt[ctr];
      start[ctr] = i;
    } else if (type == 2) {
      cin >> x;
      finish[x] = i;
    } else
      mark[i] = 1;
  }
  for (i = 1; i <= ctr; i++) {
    if (!finish[i]) finish[i] = q;
  }
  for (i = 1; i <= ctr; i++) add(1, 1, q, start[i], finish[i], i);
  solve(1, 1, q);
  for (i = 1; i <= q; i++) {
    if (mark[i]) cout << res[i] << " ";
  }
  return 0;
}
