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>
inline void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
inline void sub(long long &a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
}
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long qp(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c, b >>= 1;
  }
  return ans;
}
using namespace std;
const double eps = 1e-8;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 200000 + 10, maxn = 50000 + 10, inf = 0x3f3f3f3f;
struct ODT {
  struct node {
    int l, r;
    mutable long long val;
    node(int _l, int _r = -1, long long _val = 0) : l(_l), r(_r), val(_val) {}
    bool operator<(const node &rhs) const { return l < rhs.l; }
  };
  set<node> s;
  void ins(int l, int r, long long val) { s.insert(node(l, r, val)); }
  set<node>::iterator split(int pos) {
    set<node>::iterator p = s.lower_bound(node(pos));
    if (p != s.end() && pos == p->l) return p;
    p--;
    node te = *p;
    s.erase(p);
    s.insert(node(te.l, pos - 1, te.val));
    return s.insert(node(pos, te.r, te.val)).first;
  }
  void add(int l, int r, long long v) {
    set<node>::iterator x = split(l), y = split(r + 1);
    for (; x != y; x++) x->val += v;
  }
  void color(int l, int r, long long val) {
    set<node>::iterator x = split(l), y = split(r + 1);
    s.erase(x, y);
    s.insert(node(l, r, val));
  }
  void kth(int l, int r, long long k) {
    set<node>::iterator x = split(l), y = split(r + 1);
    vector<pair<long long, int> > v;
    for (; x != y; x++) v.push_back(make_pair(x->val, x->r - x->l + 1));
    sort(v.begin(), v.end(),
         [](pair<long long, int> a, pair<long long, int> b) {
           return a.first > b.first;
         });
    for (int i = (int)v.size() - 1; i >= 0; i--) {
      if (k <= v[i].second) {
        printf("%lld\n", v[i].first);
        return;
      }
      k -= v[i].second;
    }
  }
  void sum(int l, int r, long long a, long long b) {
    set<node>::iterator x = split(l), y = split(r + 1);
    long long ans = 0;
    for (; x != y; x++) {
      ans += 1ll * qp(x->val % b, a, 1ll * b) * (x->r - x->l + 1) % b;
      if (ans >= b) ans -= b;
    }
    printf("%lld\n", ans);
  }
  void debug() {
    for (auto x : s) printf("%d %d %d\n", x.l, x.r, x.val);
    puts("");
  }
} o;
long long n, m, seed, vmax, a[N];
long long rnd() {
  long long ret = seed;
  seed = (seed * 7ll + 13) % 1000000007;
  return ret;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
    o.ins(i, i, a[i]);
  }
  o.ins(n + 1, n + 1, 0);
  int res = 0;
  for (int i = 1; i <= m; i++) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    long long x, y;
    if (l > r) swap(l, r);
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1)
      o.add(l, r, x);
    else if (op == 2)
      o.color(l, r, x);
    else if (op == 3)
      o.kth(l, r, x);
    else
      o.sum(l, r, x, y);
  }
  return 0;
}