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 int maxn = 2e5 + 10;
struct node {
  long long l, r;
  mutable long long v;
  node(long long l, long long r, long long v) : l(l), r(r), v(v){};
  bool operator<(const node &o) const { return l < o.l; }
};
set<node> tree;
set<node>::iterator split(long long pos) {
  auto it = tree.lower_bound(node(pos, 0, 0));
  if (it != tree.end() && it->l == pos) {
    return it;
  }
  it--;
  long long l = it->l, r = it->r, v = it->v;
  tree.erase(it);
  tree.insert(node(l, pos - 1, v));
  return tree.insert(node(pos, r, v)).first;
}
int sum;
void assign(int l, int r, int v) {
  int tot = 0, len = 0;
  auto end = split(r + 1), it = split(l), begin = it;
  for (it; it != end; it++) {
    len += (it->r - it->l + 1);
    tot += it->v * (it->r - it->l + 1);
  }
  tree.erase(begin, end);
  tree.insert(node(l, r, v));
  if (v == 1)
    sum += (len - tot);
  else
    sum -= tot;
}
long long qpow(long long a, long long n, long long p) {
  long long ans = 1;
  a %= p;
  while (n) {
    if (n & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    n = n / 2;
  }
  return ans;
}
long long n, m, seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
void add(long long l, long long r, long long v) {
  auto end = split(r + 1);
  for (auto it = split(l); it != end; it++) it->v += v;
}
long long kth(long long l, long long r, long long k) {
  auto end = split(r + 1);
  vector<pair<long long, long long>> v;
  for (auto it = split(l); it != end; it++)
    v.push_back(make_pair(it->v, it->r - it->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;
  }
}
long long sum_of_pow(long long l, long long r, long long x, long long y) {
  long long tot = 0;
  auto end = split(r + 1);
  for (auto it = split(l); it != end; it++)
    tot = (tot + qpow(it->v, x, y) * (it->r - it->l + 1)) % y;
  return tot;
}
inline long long read() {
  long long ans = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans;
}
int main() {
  n = read(), m = read(), seed = read(), vmax = read();
  for (int i = 1; i <= n; ++i) {
    int r = rnd();
    tree.insert(node(i, i, r % vmax + 1));
  }
  for (int i = 1; i <= m; ++i) {
    long long opr = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    if (l > r) swap(l, r);
    if (opr == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (opr == 4) y = rnd() % vmax + 1;
    switch (opr) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        assign(l, r, x);
        break;
      case 3:
        printf("%lld\n", kth(l, r, x));
        break;
      case 4:
        printf("%lld\n", sum_of_pow(l, r, x, y));
    }
  }
  return 0;
}