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 = 1e5 + 10;
int a[MAXN];
long long n, m, seed, v;
int rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long power(long long base, int p, int mod) {
  long long ret = 1;
  base = base % mod;
  while (p) {
    if (p & 1) ret = ret * base % mod;
    p >>= 1;
    base = base * base % mod;
  }
  return ret;
}
int main() {
  cin >> n >> m >> seed >> v;
  map<int, long long> tree;
  tree[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    tree[i] = (rnd() % v) + 1;
  }
  int x, y;
  long long ret;
  for (int i = 1; i <= m; i++) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % v) + 1;
    if (op == 4) y = (rnd() % v) + 1;
    if (tree.find(l) == tree.end()) {
      tree[l] = 0;
      auto it = tree.find(l);
      it--;
      tree[l] = it->second;
    }
    if (tree.find(r + 1) == tree.end()) {
      tree[r + 1] = 0;
      auto it = tree.find(r + 1);
      it--;
      tree[r + 1] = it->second;
    }
    if (op == 1) {
      for (auto it = tree.find(l); it != tree.find(r + 1); it++) {
        it->second += x;
      }
    } else if (op == 2) {
      auto it = tree.upper_bound(l);
      while (it != tree.end() && it->first <= r) {
        tree.erase(it);
        it = tree.upper_bound(l);
      }
      tree[l] = x;
    } else if (op == 3) {
      vector<pair<long long, int> > v;
      auto it = tree.find(l);
      long long pre_val = it->second;
      int pre_idx = it->first;
      it++;
      while (true) {
        v.push_back({pre_val, it->first - pre_idx});
        if (it->first == r + 1) break;
        pre_idx = it->first;
        pre_val = it->second;
        it++;
      }
      sort(v.begin(), v.end());
      for (auto item : v) {
        if (item.second >= x) {
          ret = item.first;
          break;
        }
        x -= item.second;
      }
    } else {
      ret = 0;
      auto it = tree.find(l);
      long long pre_val = it->second;
      long long pre_idx = it->first;
      it++;
      while (true) {
        ret = (ret + (it->first - pre_idx) * power(pre_val, x, y) % y) % y;
        if (it->first == r + 1) break;
        pre_idx = it->first;
        pre_val = it->second;
        it++;
      }
    }
    if (op == 3 || op == 4) printf("%I64d\n", ret);
  }
  return 0;
}