Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,fast-math")
using namespace std;
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const int maxn = 1000 * 30 + 5;
const long long inf = 9223372036854775807;
const long long mod = 1e9 + 7;
const int lg = 20;
const long long H = 1e7 + 19;
int n, k, q, dp[lg][maxn], sum, res[maxn], cnt, st[maxn], ft[maxn], v[maxn],
    w[maxn], mark[maxn], h[4 * maxn];
vector<int> seg[4 * maxn], Q;
void upd(int l, int r, int i, int x = 1, int lx = 0, int rx = q + 1) {
  if (lx >= r || rx <= l) return;
  if (lx >= l && rx <= r) {
    seg[x].push_back(i);
    return;
  }
  int mid = lx + rx >> 1;
  upd(l, r, i, x << 1, lx, mid);
  upd(l, r, i, x << 1 | 1, mid, rx);
}
void go(int x = 1, int lx = 0, int rx = q + 1) {
  if (rx - lx == 1) return;
  h[x << 1] = h[x << 1 | 1] = h[x] + 1;
  int mid = lx + rx >> 1;
  go(x << 1, lx, mid);
  go(x << 1 | 1, mid, rx);
}
void build(int x = 1, int lx = 0, int rx = q + 1) {
  for (int i = 1; i <= k; i++) dp[h[x]][i] = dp[h[x] - 1][i];
  for (auto u : seg[x]) {
    for (int i = k; i >= 1; i--) {
      if (w[u] <= i) dp[h[x]][i] = max(dp[h[x]][i], dp[h[x]][i - w[u]] + v[u]);
    }
  }
  if (rx - lx == 1) {
    if (mark[lx]) {
      sum = 0;
      for (int i = k; i >= 1; i--) {
        sum = (long long)((long long)sum * H % mod + dp[h[x]][i]) % mod;
      }
      res[lx] = sum;
    }
    return;
  }
  int mid = lx + rx >> 1;
  build(x << 1, lx, mid);
  build(x << 1 | 1, mid, rx);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> v[i] >> w[i];
    st[i] = 0;
  }
  cnt = n;
  cin >> q;
  int t, x;
  for (int i = 1; i <= q; i++) {
    cin >> t;
    if (t == 2) {
      cin >> x;
      ft[x] = i + 1;
    }
    if (t == 1) {
      n++;
      cin >> v[n] >> w[n];
      st[n] = i;
    }
    if (t == 3) {
      Q.push_back(i);
      mark[i] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (ft[i] == 0) ft[i] = q + 1;
    upd(st[i], ft[i], i);
  }
  h[1] = 1;
  go();
  build();
  for (int u : Q) cout << res[u] << endl;
}