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 left;
  int right;
  mutable long long val;
  node(int l, int r = -1, long long v = 0) : left(l), right(r), val(v) {}
  bool operator<(const node &o) const { return left < o.left; }
};
set<node> s;
using it = set<node>::iterator;
long long qpow(long long x, long long y, long long mod) {
  long long ret = 1;
  long long base = x % mod;
  while (y) {
    if (y & 1) {
      ret = ret * base % mod;
    }
    y >>= 1;
    base = base * base % mod;
  }
  ret %= mod;
  return ret;
}
it split(int pos) {
  it ptr = s.lower_bound(node(pos));
  if (ptr != s.end() && ptr->left == pos) return ptr;
  ptr--;
  int l = ptr->left;
  int r = ptr->right;
  long long value = ptr->val;
  s.erase(ptr);
  s.insert(node(l, pos - 1, value));
  return s.insert(node(pos, r, value)).first;
}
void assign(int l, int r, long long val) {
  it itl = split(l), itr = split(r + 1);
  s.erase(itl, itr);
  s.insert(node(l, r, val));
}
void add(int l, int r, long long value) {
  it itl = split(l), itr = split(r + 1);
  for (; itl != itr; itl++) itl->val += value;
}
long long _rank(int l, int r, int k) {
  it R = split(r + 1);
  it L = split(l);
  vector<pair<long long, int> > vec;
  vec.clear();
  while (L != R) {
    vec.push_back(pair<long long, int>(L->val, L->right - L->left + 1));
    L++;
  }
  sort(vec.begin(), vec.end());
  for (auto i = vec.begin(); i != vec.end(); i++) {
    k -= i->second;
    if (k <= 0) return i->first;
  }
  return -1;
}
const long long MOD7 = 1e9 + 7;
long long sum(int L, int R, long long ex, long long mod) {
  long long ret = 0;
  it itl = split(L);
  it ir = split(R + 1);
  while (itl != ir) {
    ret += (itl->right - itl->left + 1) * qpow(itl->val, ex, mod) % mod;
    ret %= mod;
    itl++;
  }
  return ret;
}
int n, m;
long long seed;
long long vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % MOD7;
  return ret;
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) {
    s.insert(node(i, i, rnd() % vmax + 1));
  }
  s.insert(node(n + 1, n + 1, 0));
  for (int i = 1; i <= m; i++) {
    long long op = (rnd() % 4) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    long long x, y;
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1) {
      add(l, r, x);
    } else if (op == 2) {
      assign(l, r, x);
    } else if (op == 3) {
      cout << _rank(l, r, x) << '\n';
    } else {
      cout << sum(l, r, x, y) << '\n';
    }
  }
  return 0;
}