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 Nmax = 1e6 + 7;
const long long mod = 1e9 + 7LL;
struct ODT {
  struct Node {
    int l, r;
    mutable long long val;
    Node() {}
    Node(int l, int r = -1, long long val = 0) : l(l), r(r), val(val) {}
    friend bool operator<(const Node& a, const Node& b) { return a.l < b.l; }
  };
  set<Node> s;
  ODT(int n, long long x = 0) { s.insert(Node(1, n, x)); }
  inline set<Node>::iterator split(int pos) {
    auto 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 val = it->val;
    s.erase(it);
    s.insert(Node(l, pos - 1, val));
    return s.insert(Node(pos, r, val)).first;
  }
  inline void add(int l, int r, long long v = 1) {
    split(l);
    auto itr = split(r + 1), itl = split(l);
    for (; itl != itr; ++itl) itl->val += v;
  }
  inline void modify(int l, int r, long long val = 0LL) {
    split(l);
    auto itr = split(r + 1), itl = split(l);
    s.erase(itl, itr);
    s.insert(Node(l, r, val));
  }
  vector<pair<long long, int>> vp;
  inline long long rank(int l, int r, int k, bool reversed = 0) {
    if (reversed) k = r - l + 2 - k;
    split(l);
    auto itr = split(r + 1), itl = split(l);
    vp.clear();
    for (; itl != itr; ++itl) vp.push_back({itl->val, itl->r - itl->l + 1});
    sort(vp.begin(), vp.end());
    for (auto i : vp) {
      k -= i.second;
      if (k <= 0) return i.first;
    }
    return -1;
  }
  inline long long pow(long long base, long long n, long long mod) {
    base %= mod;
    long long ans = 1LL;
    while (n > 0) {
      if (n & 1) ans = ans * base % mod;
      base = base * base % mod;
      n >>= 1;
    }
    return ans;
  }
  inline long long sum(int l, int r, int ex, long long mod) {
    split(l);
    auto itr = split(r + 1), itl = split(l);
    long long ans = 0LL;
    for (; itl != itr; itl++)
      ans = (ans + ((long long)(itl->r - itl->l + 1) * pow(itl->val, ex, mod) %
                    mod)) %
            mod;
    return ans;
  }
};
int n, m;
long long seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % mod;
  return ret;
}
long long a[Nmax];
int main() {
  scanf("%d%d%lld%lld", &n, &m, &seed, &vmax);
  ODT odt(n);
  for (int i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
    odt.modify(i, i, a[i]);
  }
  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);
    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) {
      odt.add(l, r, x);
    } else if (op == 2) {
      odt.modify(l, r, x);
    } else if (op == 3) {
      printf("%lld\n", odt.rank(l, r, x));
    } else if (op == 4) {
      printf("%lld\n", odt.sum(l, r, x, y));
    }
  }
  return 0;
}