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;
using ll = long long;
const int MAXN = 1e5 + 20;
ll seed;
ll rnd() {
  ll ans = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ans;
}
ll qpow(ll base, ll exp, ll mod) {
  ll ans = 1 % mod;
  base %= mod;
  while (exp) {
    if (exp % 2) ans = ans * base % mod;
    base = base * base % mod;
    exp /= 2;
  }
  return ans;
}
class ODT {
  struct Node {
    int l, r;
    mutable ll v;
    Node(int l, int r = -1, ll v = 0) : l(l), r(r), v(v) {}
    bool operator<(Node rhs) const { return l < rhs.l; }
  };
  set<Node> s;
  int cnt = 0;
  set<Node>::iterator split(int pos);

 public:
  ODT(int n, ll *a);
  void add(int l, int r, ll x);
  void assign(int l, int r, ll x);
  ll xth(int l, int r, int x);
  ll pow(int l, int r, ll x, ll y);
};
ODT::ODT(int n, ll *a) {
  for (int i = 1; i <= n; i++) s.emplace(i, i, a[i]);
  s.emplace(n + 1, n + 1, 0);
}
set<ODT::Node>::iterator ODT::split(int pos) {
  auto it = s.lower_bound(Node(pos));
  if (it != s.end() && it->l == pos) return it;
  it--;
  int l = it->l, r = it->r;
  ll v = it->v;
  s.erase(it);
  s.emplace(l, pos - 1, v);
  return s.emplace(pos, r, v).first;
}
void ODT::add(int l, int r, ll x) {
  auto ir = split(r + 1);
  for (auto it = split(l); it != ir; it++) it->v += x;
}
void ODT::assign(int l, int r, ll x) {
  auto il = split(l), ir = split(r + 1);
  s.erase(il, ir);
  s.emplace(l, r, x);
}
ll ODT::xth(int l, int r, int x) {
  cnt++;
  vector<pair<ll, int>> vals;
  auto ir = split(r + 1);
  for (auto it = split(l); it != ir; it++)
    vals.emplace_back(it->v, it->r - it->l + 1);
  sort(vals.begin(), vals.end());
  for (auto p : vals) {
    x -= p.second;
    if (x <= 0) return p.first;
  }
  return -1;
}
ll ODT::pow(int l, int r, ll x, ll y) {
  cnt++;
  ll ans = 0;
  auto ir = split(r + 1);
  for (auto it = split(l); it != ir; it++)
    ans = (ans + (it->r - it->l + 1) * qpow(it->v, x, y)) % y;
  return ans;
}
int main() {
  int n, m;
  ll vmax;
  cin >> n >> m >> seed >> vmax;
  static ll a[MAXN];
  for (int i = 1; i <= n; i++) a[i] = rnd() % vmax + 1;
  ODT ds(n, a);
  for (int i = 1; i <= m; i++) {
    int op = rnd() % 4 + 1;
    int l = rnd() % n + 1;
    int r = rnd() % n + 1;
    if (l > r) swap(l, r);
    ll 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) ds.add(l, r, x);
    if (op == 2) ds.assign(l, r, x);
    if (op == 3) cout << ds.xth(l, r, x) << "\n";
    if (op == 4) cout << ds.pow(l, r, x, y) << "\n";
  }
}