Problem: — Willem...

— What's the matter?

— It seems that there's something wrong with Seniorious...

— I'll have a look...

<image>

Seniorious is made by linking special talismans in particular order.

After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

  * 1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
  * 2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
  * 3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It's guaranteed that 1 ≤ x ≤ r - l + 1.
  * 4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. <image>.

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:
    
    
      
    def rnd():  
      
        ret = seed  
        seed = (seed * 7 + 13) mod 1000000007  
        return ret  
      
    for i = 1 to n:  
      
        a[i] = (rnd() mod vmax) + 1  
      
    for i = 1 to m:  
      
        op = (rnd() mod 4) + 1  
        l = (rnd() mod n) + 1  
        r = (rnd() mod n) + 1  
      
        if (l > r):   
             swap(l, r)  
      
        if (op == 3):  
            x = (rnd() mod (r - l + 1)) + 1  
        else:  
            x = (rnd() mod vmax) + 1  
      
        if (op == 4):  
            y = (rnd() mod vmax) + 1  
      
    

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

Examples

Input

10 10 7 9


Output

2
1
0
3


Input

10 10 9 9


Output

1
1
3
3

Note

In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.

The operations are:

  * 2 6 7 9
  * 1 3 10 8
  * 4 4 6 2 4
  * 1 4 5 8
  * 2 1 7 1
  * 4 7 9 4 4
  * 1 2 7 9
  * 4 5 8 1 1
  * 2 5 7 5
  * 4 3 10 8 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 7;
long long n, m, seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long a[N];
long long fp(long long a, long long p, long long mod) {
  long long ans = 1, c = a % mod;
  for (long long i = 0; (1ll << i) <= p; ++i) {
    if ((p >> i) & 1) ans = (ans * c) % mod;
    c = (c * c) % mod;
  }
  return ans;
}
void cut(vector<pair<long long, long long> > &a, long long l) {
  for (auto e : a)
    if (e.first == l) return;
  long long pos = a.size();
  long long x = -1;
  for (long long i = 0; i < a.size(); ++i) {
    if (a[i].first > l) {
      pos = i;
      break;
    } else
      x = a[i].second;
  }
  a.insert(a.begin() + pos, make_pair(l, x));
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; ++i) a[i] = rnd() % vmax + 1;
  vector<pair<long long, long long> > d;
  for (long long i = 1; i <= n; ++i) d.push_back(make_pair(i, a[i]));
  long long sum = 0;
  while (m--) {
    long long op = rnd() % 4 + 1;
    long long l = rnd() % n + 1;
    long long r = rnd() % n + 1;
    if (r < l) swap(l, r);
    cut(d, l);
    if (r != n) cut(d, r + 1);
    long long x, y;
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    if (op == 1) {
      for (auto &seg : d)
        if (l <= seg.first && seg.first <= r) seg.second += x;
    } else if (op == 2) {
      vector<pair<long long, long long> > d1;
      for (auto seg : d) {
        if (l < seg.first && seg.first <= r) continue;
        if (seg.first == l)
          d1.push_back(make_pair(seg.first, x));
        else
          d1.push_back(seg);
      }
      d = d1;
    } else if (op == 3) {
      vector<pair<long long, long long> > t;
      for (long long i = 0; i < d.size(); ++i)
        if (l <= d[i].first && d[i].first <= r)
          if (i + 1 == d.size())
            t.push_back(make_pair(d[i].second, n - d[i].first + 1));
          else
            t.push_back(make_pair(d[i].second, d[i + 1].first - d[i].first));
      sort(t.begin(), t.end());
      for (auto e : t)
        if (x <= e.second) {
          cout << e.first << '\n';
          break;
        } else
          x -= e.second;
    } else {
      long long ans = 0;
      for (long long i = 0; i < d.size(); ++i)
        if (l <= d[i].first && d[i].first <= r) {
          if (i + 1 == d.size())
            ans += fp(d[i].second, x, y) * (n - d[i].first + 1);
          else
            ans += fp(d[i].second, x, y) * (d[i + 1].first - d[i].first);
        }
      cout << ans % y << '\n';
    }
  }
}