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;
template <typename T>
inline bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline void read(int &x) {
  char ch = getchar();
  int f = 1;
  x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
int n, m, seed, vmax, a[100050];
vector<pair<long long, int> > vec;
struct node {
  int l, r;
  mutable long long x;
  bool operator<(const node &x) const { return l < x.l; }
};
set<node> se;
vector<set<node>::iterator> del;
int rnd() {
  int ret = seed;
  seed = (seed * 7ll + 13) % 1000000007;
  return ret;
}
set<node>::iterator find(int x) {
  auto it = se.lower_bound(node{x, 0, 0});
  if (it != se.end() && it->l == x) return it;
  --it;
  int ll = it->l, rr = it->r;
  long long w = it->x;
  se.erase(it);
  se.insert(node{ll, x - 1, w}), se.insert(node{x, rr, w});
  return se.lower_bound(node{x, 0, 0});
}
void cov(int l, int r, int x) {
  auto fr = find(l), re = find(r + 1);
  se.erase(fr, re);
  se.insert(node{l, r, x});
}
void add(int l, int r, int x) {
  auto fr = find(l), re = find(r + 1);
  for (auto it = fr; it != re; ++it) it->x += x;
}
long long kth(int l, int r, int k) {
  vec.clear();
  auto fr = find(l), re = find(r + 1);
  for (auto it = fr; it != re; ++it)
    vec.push_back(make_pair(it->x, it->r - it->l + 1));
  sort(vec.begin(), vec.end());
  for (auto v : vec) {
    k -= v.second;
    if (k <= 0) return v.first;
  }
}
long long fpow(long long x, long long y, long long mod) {
  long long ret = 1;
  x = (x % mod + mod) % mod;
  while (y) {
    if (y & 1) ret = ret * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ret;
}
long long sum(int l, int r, int x, int y) {
  long long ret = 0;
  auto fr = find(l), re = find(r + 1);
  for (auto it = fr; it != re; ++it) {
    ret = (ret + 1ll * (it->r - it->l + 1) * fpow(it->x, x, y)) % y;
  }
  return ret;
}
int main() {
  read(n), read(m), read(seed), read(vmax);
  for (register int i = 1; i <= n; ++i) {
    a[i] = rnd() % vmax + 1;
    se.insert(node{i, i, a[i]});
  }
  se.insert(node{n + 1, n + 1, 0});
  int opt, l, r, x, y;
  while (m--) {
    opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (opt == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (opt == 4) y = rnd() % vmax + 1;
    if (opt == 1)
      add(l, r, x);
    else if (opt == 2)
      cov(l, r, x);
    else if (opt == 3)
      cout << kth(l, r, x) << endl;
    else
      cout << sum(l, r, x, y) << endl;
  }
  return 0;
}