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;
int n, m, sd, vm;
int rnd() {
  int r = sd;
  sd = (7ll * sd + 13) % 1000000007;
  return r;
}
struct node {
  int l, r;
  mutable long long v;
  node(int L, int R = 0, long long V = 0) : l(L), r(R), v(V) {}
};
bool operator<(node x, node y) { return x.l < y.l; }
set<node> s;
int po(int x, int y, int Mod) {
  int r = 1;
  while (y) {
    if (y & 1) r = 1ll * r * x % Mod;
    x = 1ll * x * x % Mod;
    y >>= 1;
  }
  return r;
}
set<node>::iterator split(int p) {
  set<node>::iterator it = s.lower_bound(node(p, 0, 0));
  if (it == s.begin()) return it;
  --it;
  int l = it->l, r = it->r;
  long long v = it->v;
  s.erase(it);
  s.insert(node(l, p - 1, v));
  return s.insert(node(p, r, v)).first;
}
void assign(int l, int r, int w) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  s.erase(itl, itr);
  s.insert(node(l, r, w));
}
void add(int l, int r, int w) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  for (; itl != itr; ++itl) itl->v += w;
}
long long qrank(int l, int r, int k) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  vector<pair<long long, int> > v;
  for (; itl != itr; ++itl) v.push_back(make_pair(itl->v, itl->r - itl->l + 1));
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); ++i) {
    k -= v[i].second;
    if (k <= 0) return v[i].first;
  }
  return -1;
}
int qpow(int l, int r, int y, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  int ans = 0;
  for (; itl != itr; ++itl)
    ans = 1ll * (ans + 1ll * po(itl->v % y, x, y) * (itl->r - itl->l + 1) % y) %
          y;
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &sd, &vm);
  for (int i = 1; i <= n; ++i) {
    int x = rnd() % vm + 1;
    s.insert(node(i, i, x));
  }
  for (int i = 1; i <= m; ++i) {
    int op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    if (l > r) swap(l, r);
    if (op == 1) add(l, r, rnd() % vm + 1);
    if (op == 2) assign(l, r, rnd() % vm + 1);
    if (op == 3) printf("%lld\n", qrank(l, r, rnd() % (r - l + 1) + 1));
    if (op == 4) printf("%d\n", qpow(l, r, rnd() % vm + 1, rnd() % vm + 1));
  }
}