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 sac = 1e9 + 7;
struct Node {
  int l, r;
  mutable long long v;
  Node(int l, int r, long long v) : l(l), r(r), v(v) {}
  bool operator<(const Node& b) const { return l < b.l; }
};
struct ODT {
  set<Node> s;
  set<Node>::iterator split(int x) {
    set<Node>::iterator it = s.lower_bound(Node(x, 0, 0));
    if (it != s.end() && it->l == x) return it;
    --it;
    int l = it->l, r = it->r;
    long long v = it->v;
    s.erase(it);
    s.insert(Node(l, x - 1, v));
    return s.insert(Node(x, r, v)).first;
  }
  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 (; itl != itr; itl++) itl->v += x;
  }
  long long kth(int l, int r, int k) {
    set<Node>::iterator itr = split(r + 1), itl = split(l);
    vector<pair<long long, int>> v;
    for (; itl != itr; itl++) v.push_back({itl->v, itl->r - itl->l + 1});
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); ++i) {
      k -= v[i].second;
      if (k <= 0) return v[i].first;
    }
  }
  int qpow(long long a, int b, int p) {
    long long res = 1;
    a %= p;
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % p;
      a = a * a % p;
    }
    return res;
  }
  long long sum(int l, int r, int x, int y) {
    set<Node>::iterator itr = split(r + 1), itl = split(l);
    long long ans = 0;
    for (; itl != itr; itl++)
      (ans += (long long)(itl->r - itl->l + 1) * qpow(itl->v, x, y)) %= y;
    return ans;
  }
} T;
int n, m, seed, vmax;
int rand() {
  int res = seed;
  seed = ((long long)seed * 7 + 13) % sac;
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; ++i) T.s.insert(Node(i, i, rand() % vmax + 1));
  for (int i = 1; i <= m; ++i) {
    int op = rand() % 4 + 1, l = rand() % n + 1, r = rand() % n + 1, x, y;
    if (l > r) swap(l, r);
    if (op == 3)
      x = rand() % (r - l + 1) + 1;
    else
      x = rand() % vmax + 1;
    if (op == 1)
      T.add(l, r, x);
    else if (op == 2)
      T.assign(l, r, x);
    else if (op == 3)
      cout << T.kth(l, r, x) << '\n';
    else {
      y = rand() % vmax + 1;
      cout << T.sum(l, r, x, y) << '\n';
    }
  }
  return 0;
}