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>
using namespace std;
const int Imx = 2147483647;
const int mod = 1000000007;
const long long Lbig = 2e18;
inline long long getnum() {
  register long long r = 0;
  register bool ng = 0;
  register char c;
  c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') ng = 1, c = getchar();
  while (c >= '0' && c <= '9') r = r * 10 + c - '0', c = getchar();
  if (ng) r = -r;
  return r;
}
template <class T>
inline void putnum(T x) {
  if (x < 0) putchar('-'), x = -x;
  register short a[20] = {}, sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
inline void putsp() { putchar(' '); }
inline void putendl() { putchar('\n'); }
inline char mygetchar() {
  register char c = getchar();
  while (c == ' ' || c == '\n') c = getchar();
  return c;
}
int n, k;
void addobject(int dp[], int v, int w) {
  for (int i = k; i >= v; i--) {
    dp[i] = max(dp[i], dp[i - v] + w);
  }
}
int type[40011], qv[40011], vol[40011], w[40011];
int st[40011], ed[40011], id[40011];
vector<int> v[200111];
void add(int x, int y, int i, int l, int r, int p) {
  if (x <= l && r <= y) {
    v[p].push_back(i);
    return;
  }
  int m = l + r >> 1;
  if (x <= m) add(x, y, i, l, m, p << 1);
  if (m < y) add(x, y, i, m + 1, r, p << 1 | 1);
}
int dp[22][1011];
void dfs(int l, int r, int p, int lv) {
  bool ok = 0;
  for (int i = l; i <= r; i++) {
    if (type[i] == 3) ok = 1;
  }
  if (!ok) return;
  for (int i = 0; i < v[p].size(); i++) {
    addobject(dp[lv], vol[v[p][i]], w[v[p][i]]);
  }
  if (l == r) {
    long long bs = 1, ans = 0;
    for (int i = 1; i <= k; i++) {
      ans += dp[lv][i] * bs % mod;
      bs = bs * 10000019 % mod;
    }
    putnum(ans % mod), putendl();
    return;
  }
  int m = l + r >> 1;
  for (int i = 0; i <= k; i++) dp[lv + 1][i] = dp[lv][i];
  dfs(l, m, p << 1, lv + 1);
  for (int i = 0; i <= k; i++) dp[lv + 1][i] = dp[lv][i];
  dfs(m + 1, r, p << 1 | 1, lv + 1);
}
int main() {
  n = getnum(), k = getnum();
  for (int i = 1; i <= n; i++) {
    type[i] = 1;
    w[i] = getnum(), vol[i] = getnum();
    qv[i] = i;
    st[i] = i;
  }
  int q;
  q = getnum();
  q += n;
  for (int i = n + 1; i <= q; i++) {
    type[i] = getnum();
    if (type[i] == 3) continue;
    qv[i] = getnum();
    if (type[i] == 1) {
      w[++n] = qv[i];
      qv[i] = n;
      vol[n] = getnum();
      st[n] = i;
    } else if (type[i] == 2) {
      ed[qv[i]] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (ed[i] == 0) ed[i] = q;
    add(st[i], ed[i], i, 1, q, 1);
  }
  dfs(1, q, 1, 0);
  return 0;
}