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;
inline int read() {
  int re = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == -1) return -1;
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    re = re * 10 + c - '0';
    c = getchar();
  }
  return re * fh;
}
inline long long readll() {
  long long re = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == -1) return -1;
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    re = re * 10 + c - '0';
    c = getchar();
  }
  return re * fh;
}
template <class T>
inline void write(T x) {
  if (x == 0) return (void)putchar('0');
  if (x < 0) putchar('-'), x = -x;
  static char buf[30];
  char* p = buf;
  while (x) {
    *(p++) = x % 10 + '0';
    x /= 10;
  }
  while (p > buf) putchar(*(--p));
}
template <>
inline void write<char*>(char* s) {
  while (*s != 0) putchar(*(s++));
}
inline int readstr(char* s) {
  char* begin = s;
  char c = getchar();
  while (c < 33 || c > 126) c = getchar();
  while (c > 32 && c < 127) {
    *(s++) = c;
    c = getchar();
  }
  *s = 0;
  return s - begin;
}
int n, m, vmax;
long long seed;
long long data() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
struct Node {
  int l;
  mutable int r;
  mutable long long v;
  Node(int l = 0, int r = 0, long long v = 0) : l(l), r(r), v(v) {}
  bool operator<(const Node& x) const { return l < x.l; }
};
int cnt;
set<Node> s;
set<Node>::iterator split(int p) {
  set<Node>::iterator it = s.lower_bound(Node(p, 0, 0));
  if (it->l == p) return it;
  --it;
  int r = it->r;
  it->r = p - 1;
  return s.insert(Node(p, r, it->v)).first;
}
void add(int l, int r, long long v) {
  set<Node>::iterator it = split(l), end = split(r + 1);
  for (; it != end; ++it) it->v += v;
}
void cover(int l, int r, long long v) {
  set<Node>::iterator it = split(l), end = split(r + 1);
  s.erase(it, end);
  s.insert(Node(l, r, v));
}
long long kth(int l, int r, int k) {
  set<Node>::iterator it = split(l), end = split(r + 1);
  vector<pair<long long, int>> v;
  for (; it != end; ++it) {
    v.push_back(pair<long long, int>(it->v, it->r - it->l + 1));
  }
  sort(v.begin(), v.end());
  for (const auto& x : v) {
    k -= x.second;
    if (k <= 0) return x.first;
  }
  return 0;
}
long long qpow(long long a, long long p, long long mod) {
  long long ret = 1, base = a % mod;
  while (p) {
    if (p & 1) ret = (ret * base) % mod;
    base = (base * base) % mod;
    p >>= 1;
  }
  return ret;
}
long long query(int l, int r, long long x, long long y) {
  set<Node>::iterator it = split(l), end = split(r + 1);
  long long ret = 0;
  for (; it != end; ++it) {
    ret =
        (ret + qpow(it->v, x, y) % y * ((long long)it->r - it->l + 1) % y) % y;
  }
  return ret;
}
int a[100100];
signed main() {
  n = read(), m = read(), seed = read(), vmax = read();
  for (int i = 1; i <= n; i++) a[i] = data() % vmax + 1;
  int p = 1, l;
  for (int p = 1; p <= n; p++) {
    l = p;
    while (a[p + 1] == a[p]) ++p;
    s.insert(Node(l, p, a[p]));
  }
  s.insert(Node(0, 0, -1));
  s.insert(Node(n + 1, n + 1, -1));
  int o, r, x, y;
  for (int i = 1; i <= m; i++) {
    o = data() % 4 + 1;
    l = data() % n + 1;
    r = data() % n + 1;
    if (l > r) swap(l, r);
    x = o == 3 ? (data() % (r - l + 1)) + 1 : (data() % vmax) + 1;
    if (o == 4)
      y = (data() % vmax) + 1;
    else
      y = -1;
    switch (o) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        cover(l, r, x);
        break;
      case 3:
        write(kth(l, r, x)), putchar('\n');
        break;
      case 4:
        write(query(l, r, x, y)), putchar('\n');
        break;
    }
  }
  return 0;
}