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 = 100007;
struct node {
  int l, r;
  mutable long long v;
};
bool operator<(const node &x, const node &y) { return x.l < y.l; }
set<node> s;
int A[maxn];
const int mod = 1e9 + 7;
long long fast_pow(long long b, int k, int md) {
  long long s = 1;
  while (k) {
    if (k & 1) s = s * b % md;
    b = b * b % md;
    k >>= 1;
  }
  return s;
}
set<node>::iterator split(int x) {
  node t;
  t.l = x;
  set<node>::iterator it = s.lower_bound(t);
  if (it != s.end() && it->l == x) return it;
  --it;
  int l = it->l, r = it->r;
  t.v = it->v;
  s.erase(it);
  t.l = l;
  t.r = x - 1;
  s.insert(t);
  t.l = x;
  t.r = r;
  return s.insert(t).first;
}
void assign(int l, int r, long long v) {
  set<node>::iterator it1 = split(l), it2 = split(r + 1);
  s.erase(it1, it2);
  s.insert((node){l, r, v});
  return;
}
int N, Q, vmax;
long long seed;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % mod;
  return ret;
}
struct pp {
  long long val;
  int sz;
};
bool operator<(const pp &x, const pp &y) { return x.val < y.val; }
vector<pp> d;
long long kth(int l, int r, int k) {
  set<node>::iterator it1 = split(l), it2 = split(r + 1);
  d.clear();
  while (it1 != it2) {
    d.push_back((pp){it1->v, it1->r - it1->l + 1});
    ++it1;
  }
  sort(d.begin(), d.end());
  for (int i = 0; i < (int)d.size(); i++)
    if (d[i].sz >= k)
      return d[i].val;
    else
      k -= d[i].sz;
  return -1;
}
long long sum(int l, int r, int k, int md) {
  set<node>::iterator it1 = split(l), it2 = split(r + 1);
  long long res = 0;
  while (it1 != it2) {
    res += fast_pow(it1->v % md, k, md) * (long long)(it1->r - it1->l + 1) % md;
    res %= md;
    ++it1;
  }
  return res;
}
void add(int l, int r, int k) {
  set<node>::iterator it1 = split(l), it2 = split(r + 1);
  while (it1 != it2) it1->v += k, ++it1;
  return;
}
int main() {
  scanf("%d%d%lld%d", &N, &Q, &seed, &vmax);
  int op, l, r, x;
  node t;
  for (int i = 1; i <= N; i++) {
    A[i] = rnd() % vmax + 1;
    t.l = t.r = i;
    t.v = A[i];
    s.insert(t);
  }
  t.l = t.r = N + 1;
  t.v = 0;
  s.insert(t);
  while (Q--) {
    op = rnd() % 4 + 1;
    l = rnd() % N + 1;
    r = rnd() % N + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      x = rnd() % vmax + 1;
      add(l, r, x);
    } else if (op == 2) {
      x = rnd() % vmax + 1;
      assign(l, r, x);
    } else if (op == 3) {
      x = rnd() % (r - l + 1) + 1;
      printf("%lld\n", kth(l, r, x));
    } else if (op == 4) {
      x = rnd() % vmax + 1;
      int y = rnd() % vmax + 1;
      printf("%lld\n", sum(l, r, x, y));
    }
  }
  return 0;
}