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;
int n, m, vm, M = 1000000007, fin[100005], op, l, r, x, y;
long long seed, a[100005];
set<int> st;
int rnd() {
  int res = seed;
  seed = (seed * 7 + 13) % M;
  return res;
}
long long pot(long long b, int p, int m) {
  long long res = 1;
  b %= m;
  while (p) {
    if (p & 1) {
      res *= b;
      res %= m;
    }
    b *= b;
    b %= m;
    p >>= 1;
  }
  return res;
}
void sumOp() {
  auto it = st.upper_bound(l);
  it--;
  if (*it < l && fin[*it] >= l) {
    fin[l] = fin[*it];
    fin[*it] = l - 1;
    a[l] = a[*it];
    st.insert(l);
    it++;
  } else if (fin[*it] < l) {
    it++;
  }
  while (it != st.end() && fin[*it] <= r) {
    a[*it] += x;
    it++;
  }
  if (it != st.end() && *it <= r) {
    a[r + 1] = a[*it];
    a[*it] += x;
    st.insert(r + 1);
    fin[r + 1] = fin[*it];
    fin[*it] = r;
  }
}
void update() {
  auto it = st.upper_bound(l);
  it--;
  if (*it < l) {
    fin[l] = fin[*it];
    fin[*it] = l - 1;
    a[l] = x;
    st.insert(l);
    if (fin[l] > r) {
      fin[r + 1] = fin[l];
      st.insert(r + 1);
      a[r + 1] = a[*it];
    }
    it++;
  } else if (*it == l) {
    if (fin[l] > r) {
      fin[r + 1] = fin[l];
      st.insert(r + 1);
      a[r + 1] = a[*it];
    }
  }
  a[l] = x;
  while (it != st.end() && *it <= l) it++;
  while (it != st.end() && fin[*it] <= r) {
    auto aux = it;
    it++;
    st.erase(aux);
  }
  if (it != st.end() && *it <= r) {
    a[r + 1] = a[*it];
    st.insert(r + 1);
    fin[r + 1] = fin[*it];
    st.erase(it);
    it++;
  }
  fin[l] = r;
}
long long xMN() {
  auto it = st.upper_bound(l);
  it--;
  priority_queue<pair<long long, int> > pq;
  int acu = 0;
  if (*it < l) {
    acu += min(fin[*it], r) - l + 1;
    pq.emplace(a[*it], min(fin[*it], r) - l + 1);
    it++;
  }
  while (it != st.end() && fin[*it] <= r) {
    acu += fin[*it] - *it + 1;
    pq.emplace(a[*it], fin[*it] - *it + 1);
    it++;
  }
  if (it != st.end() && *it <= r) {
    acu += r - *it + 1;
    pq.emplace(a[*it], r - *it + 1);
  }
  while (acu - pq.top().second >= x) {
    acu -= pq.top().second;
    pq.pop();
  }
  return pq.top().first;
}
int obtRes() {
  auto it = st.upper_bound(l);
  it--;
  long long sum = 0;
  if (*it < l) {
    sum += pot(a[*it], x, y) * (min(fin[*it], r) - l + 1);
    sum %= y;
    it++;
  }
  while (it != st.end() && fin[*it] <= r) {
    sum += pot(a[*it], x, y) * (fin[*it] - *it + 1);
    sum %= y;
    it++;
  }
  if (it != st.end() && *it <= r) {
    sum += pot(a[*it], x, y) * (r - *it + 1);
    sum %= y;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> seed >> vm;
  for (int i = 1; i <= n; i++) {
    a[i] = rnd() % vm + 1;
    st.insert(i);
    fin[i] = i;
  }
  while (m--) {
    op = rnd() % 4;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 0) {
      x = rnd() % vm + 1;
      sumOp();
    } else if (op == 1) {
      x = rnd() % vm + 1;
      update();
    } else if (op == 2) {
      x = rnd() % (r - l + 1) + 1;
      cout << xMN() << '\n';
    } else {
      x = rnd() % vm + 1;
      y = rnd() % vm + 1;
      cout << obtRes() << '\n';
    }
  }
  return 0;
}