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 N = 100002;
const long long MX = 1e14 + 1e9 + 2;
int n, m, seed, vmax;
long long tree[4 * N];
vector<pair<long long, long long> > a;
int rnd() {
  int ret = seed;
  seed = (seed * 7ll + 13) % 1000000007;
  return ret;
}
long long power(int x, int y, int M) {
  if (!y) return 1;
  long long sq = power(x, y / 2, M);
  sq = sq * sq % M;
  if (y & 1) sq = sq * x % M;
  return sq;
}
inline void build(int i, int l, int r) {
  if (l == r) {
    tree[i] = (rnd() % vmax) + 1;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * i, l, mid);
  build(2 * i + 1, mid + 1, r);
  if (tree[2 * i] == tree[2 * i + 1]) tree[i] = tree[2 * i];
}
void query(int i, int start, int end, int l, int r) {
  if (tree[i] and start != end) {
    tree[2 * i] = tree[i];
    tree[2 * i + 1] = tree[i];
  }
  if (l <= start && r >= end and tree[i]) {
    a.push_back(pair<long long, long long>(tree[i], end - start + 1));
    return;
  }
  if (r < start || l > end) return;
  int mid = (start + end) / 2;
  query(2 * i, start, mid, l, r);
  query(2 * i + 1, mid + 1, end, l, r);
}
long long solve(int i, int start, int end, int l, int r, int x, int y) {
  if (tree[i] and start != end) {
    tree[2 * i] = tree[i];
    tree[2 * i + 1] = tree[i];
  }
  if (l <= start && r >= end and tree[i])
    return power(tree[i] % y, x, y) * (end - start + 1);
  if (r < start || l > end) return 0;
  int mid = (start + end) / 2;
  return solve(2 * i, start, mid, l, r, x, y) +
         solve(2 * i + 1, mid + 1, end, l, r, x, y);
}
void update(int i, int start, int end, int l, int r, int v) {
  if (tree[i] and start != end) {
    tree[2 * i] = tree[i];
    tree[2 * i + 1] = tree[i];
  }
  if (l <= start && r >= end and tree[i]) {
    tree[i] += v;
    return;
  }
  if (r < start || l > end) return;
  int mid = (start + end) / 2;
  update(2 * i, start, mid, l, r, v);
  update(2 * i + 1, mid + 1, end, l, r, v);
  if (tree[2 * i] == tree[2 * i + 1])
    tree[i] = tree[2 * i];
  else
    tree[i] = 0;
}
void Set(int i, int start, int end, int l, int r, int v) {
  if (tree[i] and start != end) {
    tree[2 * i] = tree[i];
    tree[2 * i + 1] = tree[i];
  }
  if (l <= start && r >= end) {
    tree[i] = v;
    return;
  }
  if (r < start || l > end) return;
  int mid = (start + end) / 2;
  Set(2 * i, start, mid, l, r, v);
  Set(2 * i + 1, mid + 1, end, l, r, v);
  if (tree[2 * i] == tree[2 * i + 1])
    tree[i] = tree[2 * i];
  else
    tree[i] = 0;
}
inline void generate(int &op, int &l, int &r, int &x, int &y) {
  op = (rnd() % 4) + 1;
  l = (rnd() % n) + 1;
  r = (rnd() % n) + 1;
  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;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%d%d", &seed, &vmax);
  build(1, 1, n);
  while (m--) {
    int op, l, r, x, y;
    generate(op, l, r, x, y);
    if (op == 1)
      update(1, 1, n, l, r, x);
    else if (op == 2)
      Set(1, 1, n, l, r, x);
    else if (op == 3) {
      a.clear();
      query(1, 1, n, l, r);
      sort(a.begin(), a.end());
      for (pair<long long, long long> p : a) {
        if (p.second >= x) {
          printf("%I64d\n", p.first);
          break;
        }
        x -= p.second;
      }
    } else
      printf("%I64d\n", solve(1, 1, n, l, r, x, y) % y);
  }
}