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 long long maxn = 40100, mod = 1e9 + 7, maxa = 1100, maxb = 22, base = 737,
                base2 = 3079, mod2 = 1e7 + 19;
const long long inf = 2e18 + 13;
long long max(long long x, long long y) { return (x > y ? x : y); }
long long min(long long x, long long y) { return (x < y ? x : y); }
vector<long long> seg[4 * maxn];
long long dp[maxb][maxa];
bool query[maxn];
long long in[maxn];
long long out[maxn];
long long w[maxn];
long long v[maxn];
long long n, k;
void update(long long x, long long l, long long r, long long s, long long t,
            long long val) {
  if (s <= l && r <= t) {
    seg[x].push_back(val);
    return;
  }
  long long mid = (l + r) / 2;
  if (s < mid) {
    update(2 * x, l, mid, s, min(mid, t), val);
  }
  if (t > mid) {
    update(2 * x + 1, mid, r, max(mid, s), t, val);
  }
}
void redo(long long h) {
  for (long long j = 0; j < maxa; j++) {
    dp[h + 1][j] = dp[h][j];
  }
}
void dfs(long long x, long long l, long long r, long long h) {
  for (auto ind : seg[x]) {
    for (long long j = maxa - 1; j >= w[ind]; j--) {
      dp[h][j] = max(dp[h][j], dp[h][j - w[ind]] + v[ind]);
    }
  }
  if (r - l == 1) {
    if (!query[l]) {
      return;
    }
    long long mx = -inf;
    long long ans = 0;
    long long pw = 1;
    for (long long j = 1; j <= k; j++) {
      mx = max(mx, dp[h][j]);
      ans += mx * pw % mod;
      ans %= mod;
      pw *= mod2;
      pw %= mod;
    }
    cout << ans << endl;
    return;
  }
  long long mid = (l + r) / 2;
  redo(h);
  dfs(2 * x, l, mid, h + 1);
  redo(h);
  dfs(2 * x + 1, mid, r, h + 1);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> v[i] >> w[i];
    in[i] = 0;
    out[i] = maxn;
  }
  long long q;
  long long pt = n;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      cin >> v[pt] >> w[pt];
      in[pt] = i;
      out[pt] = maxn;
      pt++;
      continue;
    }
    if (t == 2) {
      long long ind;
      cin >> ind;
      ind--;
      out[ind] = i;
      continue;
    }
    query[i] = 1;
  }
  for (long long i = 0; i < pt; i++) {
    update(1, 0, maxn, in[i], out[i], i);
  }
  dfs(1, 0, maxn, 0);
  return 0;
}
