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 = 1e5 + 7;
const long long mod = 1e9 + 7;
struct node {
  int l, r;
  mutable long long val;
  bool operator<(const node& B) const { return l < B.l; }
  node(int _l, int _r, long long _val) : l(_l), r(_r), val(_val) {}
  node(int _l) : l(_l) {}
};
set<node> Tree;
set<node>::iterator Split(int pos) {
  set<node>::iterator it = Tree.lower_bound(node(pos));
  if (it != Tree.end() && it->l == pos) return it;
  --it;
  int l = it->l, r = it->r;
  long long val = it->val;
  Tree.erase(it);
  Tree.insert(node(l, pos - 1, val));
  return Tree.insert(node(pos, r, val)).first;
}
void Assign(int l, int r, long long x) {
  set<node>::iterator R = Split(r + 1), L = Split(l);
  Tree.erase(L, R);
  Tree.insert(node(l, r, x));
}
void Add(int l, int r, long long x) {
  set<node>::iterator R = Split(r + 1), L = Split(l);
  while (L != R) {
    L->val += x;
    ++L;
  }
}
int n, m;
long long Seed, Vmax, V[maxn];
long long Rand() {
  long long tmp = Seed;
  Seed = (Seed * 7 + 13) % mod;
  return tmp;
}
long long qpow(long long x, long long c, const long long md) {
  x %= md;
  long long res = 1;
  while (c) {
    if (c & 1) res = (res * x) % md;
    c >>= 1;
    x = (x * x) % md;
  }
  return res;
}
struct NODE {
  long long x, cnt;
  bool operator<(const NODE& B) const { return x < B.x; }
};
vector<NODE> vec;
void KTH(int l, int r, long long k) {
  vec.clear();
  set<node>::iterator pos2 = Split(r + 1), pos1 = Split(l);
  while (pos1 != pos2) {
    vec.push_back((NODE){pos1->val, pos1->r - pos1->l + 1});
    ++pos1;
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); ++i) {
    k -= vec[i].cnt;
    if (k <= 0) {
      printf("%lld\n", vec[i].x);
      return;
    }
  }
}
void POW(int l, int r, long long x, const long long y) {
  long long res = 0;
  set<node>::iterator pos2 = Split(r + 1), pos1 = Split(l);
  while (pos1 != pos2) {
    res = (res + (pos1->r - pos1->l + 1) % y * qpow(pos1->val, x, y) % y) % y;
    ++pos1;
  }
  printf("%lld\n", res);
}
int main() {
  scanf("%d %d %lld %lld", &n, &m, &Seed, &Vmax);
  int l = 1;
  long long last;
  for (int i = 1; i <= n; ++i) {
    V[i] = Rand() % Vmax + 1;
    if (i == 1) {
      last = V[i];
    } else {
      if (last != V[i]) {
        Tree.insert(node(l, i - 1, last));
        last = V[i];
        l = i;
      }
    }
    if (i == n) Tree.insert(node(l, i, last));
  }
  while (m--) {
    int op = Rand() % 4 + 1;
    l = Rand() % n + 1;
    int r = Rand() % n + 1;
    long long x, y;
    if (l > r) swap(l, r);
    if (op == 3) {
      x = Rand() % (r - l + 1) + 1;
    } else {
      x = Rand() % Vmax + 1;
    }
    if (op == 4) {
      y = Rand() % Vmax + 1;
    }
    if (op == 1) {
      Add(l, r, x);
    }
    if (op == 2) {
      Assign(l, r, x);
    }
    if (op == 3) {
      KTH(l, r, x);
    }
    if (op == 4) {
      POW(l, r, x, y);
    }
  }
  return 0;
}