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 f1 = 10000019;
const int mod = 1000000007;
struct pi {
  vector<pair<int, int> > g;
  int le, ri;
} pp[maxn << 2];
int f[maxn], e[maxn];
void init(int tot, int l, int r) {
  pp[tot].le = l;
  pp[tot].ri = r;
  pp[tot].g.clear();
  if (l == r) return;
  init(2 * tot, l, (l + r) / 2);
  init(2 * tot + 1, (l + r) / 2 + 1, r);
}
void add(int tot, int l, int r, pair<int, int> p2) {
  if (l > r) return;
  if (l <= pp[tot].le && pp[tot].ri <= r) {
    pp[tot].g.push_back(p2);
    return;
  }
  int mid = (pp[tot].le + pp[tot].ri) / 2;
  if (l <= mid) add(2 * tot, l, r, p2);
  if (r > mid) add(2 * tot + 1, l, r, p2);
}
int ss = 0;
int ans[maxn];
int dp[20][1005];
int k;
void update(int le, pair<int, int> p2) {
  int i;
  for (i = k; i >= p2.second; i--) {
    dp[le][i] = max(dp[le][i], dp[le][i - p2.second] + p2.first);
  }
}
void solve(int le, int tot) {
  int i;
  for (i = 0; i < pp[tot].g.size(); i++) {
    update(le, pp[tot].g[i]);
  }
  if (pp[tot].le == pp[tot].ri) {
    long long s = 0;
    for (i = k; i >= 1; i--) {
      s = s * f1 + dp[le][i];
      s %= mod;
    }
    ans[++ss] = s;
    return;
  }
  for (i = 0; i <= k; i++) {
    dp[le + 1][i] = dp[le][i];
  }
  solve(le + 1, 2 * tot);
  for (i = 0; i <= k; i++) {
    dp[le + 1][i] = dp[le][i];
  }
  solve(le + 1, 2 * tot + 1);
}
vector<pair<int, int> > g;
int main() {
  int i, n, m, cnt = 0;
  m = 0;
  cin >> n >> k;
  memset(e, -1, sizeof(e));
  for (i = 1; i <= n; i++) {
    f[i] = cnt + 1;
    int a, b;
    scanf("%d%d", &a, &b);
    g.push_back(make_pair(a, b));
  }
  cin >> m;
  for (i = 0; i < m; i++) {
    int p;
    scanf("%d", &p);
    if (p == 3) {
      cnt++;
    } else if (p == 1) {
      n++;
      f[n] = cnt + 1;
      int a, b;
      scanf("%d%d", &a, &b);
      g.push_back(make_pair(a, b));
    } else {
      int p;
      scanf("%d", &p);
      e[p] = cnt;
    }
  }
  init(1, 1, cnt);
  for (i = 0; i < g.size(); i++) {
    if (e[i + 1] == -1) {
      e[i + 1] = cnt;
    }
    add(1, f[i + 1], e[i + 1], g[i]);
  }
  solve(0, 1);
  for (i = 1; i <= cnt; i++) {
    printf("%d\n", ans[i]);
  }
}
