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>
const int MAXN = 100000;
const int MOD = 1000000007;
using namespace std;
inline int read() {
  int x = 0, w = 1;
  char c = ' ';
  while (c < '0' || c > '9') {
    c = getchar();
    if (c == '-') w = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * w;
}
long long ksm(int x, int k, long long m) {
  if (!k) return 1;
  long long tmp = ksm(x, k / 2, m) % m;
  if (k % 2)
    return tmp * tmp % m * x % m;
  else
    return tmp * tmp % m;
}
struct ChthollyTree {
  struct Node {
    int l, r;
    mutable long long v;
    bool operator<(const Node &rhs) const { return l < rhs.l; }
    Node(int _l, int _r = -1, long long _v = 0) {
      l = _l;
      r = _r;
      v = _v;
    }
  };
  set<Node> S;
  vector<pair<long long, int> > V;
  set<Node>::iterator split(int pos) {
    set<Node>::iterator it = S.lower_bound(Node(pos));
    if (it != S.end() && it->l == pos) return it;
    --it;
    if (pos > it->r) return S.end();
    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;
  }
  void add(int l, int r, int num) {
    split(l);
    set<Node>::iterator itr = split(r + 1), itl = split(l);
    for (set<Node>::iterator it = itl; it != itr; ++it) it->v += num;
  }
  void update(int l, int r, int num) {
    split(l);
    set<Node>::iterator itr = split(r + 1), itl = split(l);
    S.erase(itl, itr);
    S.insert(Node(l, r, num));
  }
  long long getRank(int l, int r, int k) {
    V.clear();
    split(l);
    set<Node>::iterator itr = split(r + 1), itl = split(l);
    for (set<Node>::iterator it = itl; it != itr; ++it)
      V.push_back(make_pair(it->v, it->r - it->l + 1));
    sort(V.begin(), V.end());
    vector<pair<long long, int> >::iterator it;
    for (it = V.begin(); it != V.end(); ++it) {
      k -= it->second;
      if (k <= 0) return it->first;
    }
    return -1LL;
  }
  long long query(int l, int r, int x, long long mod) {
    split(l);
    set<Node>::iterator itr = split(r + 1), itl = split(l);
    long long ret = 0;
    for (set<Node>::iterator it = itl; it != itr; ++it)
      ret = (ret % mod + (long long)(it->r - it->l + 1) % mod *
                             ksm(it->v % mod, x, mod) % mod) %
            mod;
    return ret;
  }
  void insert(int l, int r, int num) { S.insert(Node(l, r, num)); }
  void debug() {
    for (set<Node>::iterator it = S.begin(); it != S.end(); ++it)
      cout << it->l << ' ' << it->r << ' ' << it->v << endl;
    cout << endl;
  }
} C;
int n, m, vmax;
int a[MAXN + 5];
long long seed;
int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % MOD;
  return ret;
}
void init() {
  n = read();
  m = read();
  seed = read();
  vmax = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = rnd() % vmax + 1;
    C.insert(i, i, a[i]);
  }
}
int main() {
  init();
  for (int i = 1; i <= m; ++i) {
    int opt = rnd() % 4 + 1;
    int l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    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;
    switch (opt) {
      case 1:
        C.add(l, r, x);
        break;
      case 2:
        C.update(l, r, x);
        break;
      case 3:
        printf("%I64d\n", C.getRank(l, r, x));
        break;
      case 4:
        printf("%I64d\n", C.query(l, r, x, y));
        break;
    }
  }
  return 0;
}