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;
using ll = long long;
struct OldDriverTree {
  using dtype = ll;
  struct Node {
    int l, r;
    mutable dtype v;
    inline bool operator<(const Node &ano) const { return l < ano.l; }
  };
  using Iter = set<Node>::iterator;
  using Range = pair<Iter, Iter>;
  set<Node> segs;
  inline Iter find(int x) { return --segs.upper_bound({x, 0, 0}); }
  void init(int a[], int n) {
    segs.clear();
    Iter hint = segs.begin();
    for (int l = 0, r; l < n; l = r) {
      r = l + 1;
      while (r < n && a[r] == a[l]) r++;
      segs.insert(hint, {l, r, a[l]});
    }
  }
  Iter split(int x) {
    Iter iter = find(x);
    if (!(x < iter->r)) return segs.end();
    if (!(x > iter->l)) return iter;
    Node seg = *iter;
    Iter hint = segs.erase(iter);
    hint = segs.insert(hint, {seg.l, x, seg.v});
    return segs.insert(hint, {x, seg.r, seg.v});
  }
  Range get_range(int l, int r) {
    split(l);
    Iter end = split(r);
    return Range(find(l), end);
  }
  void assign(int l, int r, int v) {
    Range range = get_range(l, r);
    Iter hint = segs.erase(range.first, range.second);
    hint = segs.insert(hint, {l, r, v});
  }
  void seg_add(int l, int r, int v) {
    Range range = get_range(l, r);
    for (Iter i = range.first; i != range.second; i++) {
      i->v += v;
    }
  }
  dtype kth(int l, int r, int k) {
    vector<pair<dtype, int>> list;
    Range range = get_range(l, r);
    for (Iter i = range.first; i != range.second; i++) {
      list.emplace_back(i->v, i->r - i->l);
    }
    sort(list.begin(), list.end());
    for (auto p : list) {
      if (!(p.second >= k)) {
        k -= p.second;
        continue;
      }
      return p.first;
    }
    assert(false);
  }
  inline int add(int a, int b, int mod) {
    if ((a += b) >= mod) a -= mod;
    return a;
  }
  inline int mul(int a, int b, int mod) { return a * 1ll * b % mod; }
  int qpow(int a, int b, int mod) {
    int r = 1;
    for (; b; b >>= 1) {
      if (b & 1) r = mul(r, a, mod);
      a = mul(a, a, mod);
    }
    return r;
  }
  int pow_sum(int l, int r, int x, int y) {
    Range range = get_range(l, r);
    int res = 0;
    for (Iter i = range.first; i != range.second; i++) {
      int buf = qpow(i->v % y, x, y);
      buf = mul(buf, i->r - i->l, y);
      res = add(res, buf, y);
    }
    return res;
  }
  void print() {
    cout << "\"===odt begin===\""
         << " = " << ("===odt begin===") << "\n";
    for (Node nd : segs) {
      cout << "nd.l"
           << " = " << (nd.l) << ", ";
      cout << "nd.r"
           << " = " << (nd.r) << ", ";
      cout << "nd.v"
           << " = " << (nd.v) << "\n";
    }
    cout << "\"===odt end=====\\n\""
         << " = " << ("===odt end=====\n") << "\n";
  }
} odt;
const int N = 1e5 + 7;
int n, m, vmax, seed;
int a[N], l[N], r[N], x[N], y[N], op[N];
struct Random {
  const int MOD = 1e9 + 7;
  int add(int a, int b) {
    if ((a += b) >= MOD) a -= MOD;
    return a;
  }
  int mul(int a, int b) { return a * 1ll * b % MOD; }
  int rnd() {
    int ret = seed;
    seed = add(mul(seed, 7), 13);
    return ret;
  }
  void build() {
    for (int i = 0; i < n; i++) a[i] = (rnd() % vmax) + 1;
    for (int i = 0; i < m; i++) {
      op[i] = (rnd() % 4) + 1;
      l[i] = (rnd() % n) + 1;
      r[i] = (rnd() % n) + 1;
      if (l[i] > r[i]) swap(l[i], r[i]);
      if (op[i] == 3)
        x[i] = (rnd() % (r[i] - l[i] + 1)) + 1;
      else
        x[i] = (rnd() % vmax) + 1;
      if (op[i] == 4) y[i] = (rnd() % vmax) + 1;
    }
  }
} ran;
void solve() {
  cin >> n >> m >> seed >> vmax;
  ran.build();
  odt.init(a, n);
  for (int i = 0; i < m; i++) {
    l[i]--;
    switch (op[i]) {
      case 1:
        odt.seg_add(l[i], r[i], x[i]);
        break;
      case 2:
        odt.assign(l[i], r[i], x[i]);
        break;
      case 3:
        cout << odt.kth(l[i], r[i], x[i]) << "\n";
        break;
      case 4:
        cout << odt.pow_sum(l[i], r[i], x[i], y[i]) << "\n";
        break;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}