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;
struct node {
  int l;
  int r;
  mutable long long int v;
  bool operator<(const node &a) const { return l < a.l; }
};
int n, m;
long long int seed, maxv;
set<node> tree;
long long int rnd() {
  long long int ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long int poww(long long int v, long long int bit, long long int mod) {
  long long int res = 1;
  long long int out = v % mod;
  while (bit) {
    if (bit & 1) {
      res = res * out % mod;
    }
    out = out * out % mod;
    bit >>= 1;
  }
  return res;
}
set<node>::iterator split(int pos) {
  set<node>::iterator iter = tree.lower_bound((node){pos, 0, 0});
  if (iter != tree.end() && iter->l == pos) {
    return iter;
  }
  iter--;
  int ll = iter->l, rr = iter->r;
  long long int v = iter->v;
  tree.erase(iter);
  tree.insert((node){ll, pos - 1, v});
  return tree.insert((node){pos, rr, v}).first;
}
void assign(int l, int r, int v) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  tree.erase(itl, itr);
  tree.insert((node){l, r, v});
  return;
}
void add(int l, int r, long long int val) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; itl++) {
    itl->v += val;
  }
  return;
}
long long int rk(int l, int r, int k) {
  vector<pair<long long int, int> > ve;
  set<node>::iterator itr = split(r + 1), itl = split(l);
  ve.clear();
  for (; itl != itr; itl++) {
    ve.push_back(pair<long long int, int>(itl->v, itl->r - itl->l + 1));
  }
  sort(ve.begin(), ve.end());
  for (vector<pair<long long int, int> >::iterator it = ve.begin();
       it != ve.end(); it++) {
    k -= it->second;
    if (k <= 0) {
      return it->first;
    }
  }
  return (long long int)-1;
}
long long int sum(int l, int r, int bit, int mod) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  long long int res = 0;
  for (; itl != itr; itl++) {
    res =
        (res + (long long int)(itl->r - itl->l + 1) * poww(itl->v, bit, mod)) %
        mod;
  }
  return res;
}
int main(void) {
  scanf("%d%d%lld%lld", &n, &m, &seed, &maxv);
  for (int i = 1; i <= n; i++) {
    tree.insert((node){i, i, rnd() % maxv + 1});
  }
  int order, l, r, x, y;
  for (int i = 0; i < m; i++) {
    order = rnd() % 4 + 1;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) {
      swap(l, r);
    }
    switch (order) {
      case 1:
        x = rnd() % maxv + 1;
        add(l, r, x);
        break;
      case 2:
        x = rnd() % maxv + 1;
        assign(l, r, x);
        break;
      case 3:
        x = rnd() % (r - l + 1) + 1;
        printf("%lli\n", rk(l, r, x));
        break;
      case 4:
        x = rnd() % maxv + 1;
        y = rnd() % maxv + 1;
        printf("%lli\n", sum(l, r, x, y));
        break;
    }
  }
  return 0;
}