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 read() {
  char ch = getchar();
  int x = 0, pd = 0;
  while (ch < '0' || ch > '9') pd ^= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return pd ? -x : x;
}
const int maxn = 100005;
const int mod = 1000000007;
int n, m, vmax;
long long seed;
int rnd() {
  int res = seed;
  seed = (seed * 7 + 13) % mod;
  return res;
}
struct node {
  int l, r;
  mutable long long v;
  node(int L = 0, int R = 0, long long val = 0) : l(L), r(R), v(val) {}
  bool operator<(node x) const { return l < x.l; }
};
set<node> s;
set<node>::iterator split(int pos) {
  if (pos == n + 1) return s.end();
  set<node>::iterator it = s.lower_bound(node(pos));
  if (it != s.end() && it->l == pos) return it;
  --it;
  int l = it->l, r = it->r;
  long long v = it->v;
  s.erase(it);
  s.insert(node(l, pos - 1, v));
  return s.insert(node(pos, r, v)).first;
}
node a[maxn];
bool cmp(node x, node y) { return x.v < y.v; }
long long kth(int l, int r, int k) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  int len = 0;
  for (set<node>::iterator it = itl; it != itr; ++it) a[++len] = (*it);
  sort(a + 1, a + len + 1, cmp);
  long long res = 0;
  for (int i = 1; i <= len; i++)
    if (k <= a[i].r - a[i].l + 1) {
      res = a[i].v;
      break;
    } else
      k -= a[i].r - a[i].l + 1;
  return res;
}
int p;
int fastpow(int x, int y) {
  int res = 1;
  for (; y; x = 1ll * x * x % p, y >>= 1)
    if (y & 1) res = 1ll * res * x % p;
  return res;
}
int query(int l, int r, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  int res = 0;
  for (set<node>::iterator it = itl; it != itr; ++it)
    res = (1ll * fastpow((it->v) % p, x) * (it->r - it->l + 1) % p + res) % p;
  return res;
}
void assign(int l, int r, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  s.erase(itl, itr);
  s.insert(node(l, r, x));
}
void add(int l, int r, int x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  for (set<node>::iterator it = itl; it != itr; ++it) (it->v) += x;
}
int main() {
  n = read(), m = read(), seed = read(), vmax = read();
  for (int i = 1; i <= n; i++) s.insert(node(i, i, rnd() % vmax + 1));
  while (m--) {
    int opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    int x;
    if (opt == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (opt == 4)
      p = rnd() % vmax + 1, printf("%d\n", query(l, r, x));
    else if (opt == 3)
      printf("%lld\n", kth(l, r, x));
    else if (opt == 2)
      assign(l, r, x);
    else
      add(l, r, x);
  }
  return 0;
}