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;
long long mod = 1000000007;
const long long inf = (1ll << 60) + 5;
long long n, m, seed, vmax;
long long a[100009];
struct node {
  long long l, r;
  long long val;
  node(long long L, long long R, long long V) : l(L), r(R), val(V) {}
  node() {}
  friend bool operator<(node a, node b) {
    if (a.l == b.l) return a.r < b.r;
    return a.l < b.l;
  }
};
set<node> S;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7LL + 13LL) % 1000000007;
  return ret;
}
long long pow(long long a, long long b, long long c) {
  long long r = 1;
  a = a % c;
  while (b) {
    if (b & 1) r = r * a % c;
    a = a * a % c;
    b >>= 1;
  }
  return r % c;
}
void cut(long long l) {
  node temp(l, inf, inf);
  set<node>::iterator it = S.upper_bound(temp);
  it--;
  if (it->l != l) {
    node p1, p2;
    p1.l = it->l;
    p1.r = l - 1;
    p1.val = it->val;
    p2.l = l;
    p2.r = it->r;
    p2.val = it->val;
    S.erase(it);
    S.insert(p1);
    S.insert(p2);
  }
}
void op1(long long l, long long r, long long x) {
  cut(l);
  if (r + 1 <= n) cut(r + 1);
  node temp(l, inf, inf);
  set<node>::iterator it = S.upper_bound(temp);
  set<node>::iterator tmp;
  it--;
  vector<node> List;
  while (it != S.end()) {
    if (it->r > r) break;
    tmp = it;
    node p = (*it);
    it++;
    p.val += x;
    S.erase(tmp);
    List.push_back(p);
  }
  for (auto i : List) {
    S.insert(i);
  }
}
void op2(long long l, long long r, long long x) {
  cut(l);
  if (r + 1 <= n) cut(r + 1);
  node temp(l, inf, inf);
  set<node>::iterator it = S.upper_bound(temp);
  set<node>::iterator tmp;
  it--;
  while (it != S.end()) {
    if (it->r > r) break;
    tmp = it;
    it++;
    S.erase(tmp);
  }
  S.insert(node(l, r, x));
}
bool cmp(node a, node b) { return a.val < b.val; }
long long op3(long long l, long long r, long long x) {
  cut(l);
  if (r + 1 <= n) cut(r + 1);
  node temp(l, inf, inf);
  set<node>::iterator it = S.upper_bound(temp);
  set<node>::iterator tmp;
  it--;
  vector<node> ans;
  while (it != S.end()) {
    if (it->r > r) break;
    node t = *it;
    it++;
    ans.push_back(t);
  }
  sort(ans.begin(), ans.end(), cmp);
  for (int i = 0; i < ans.size(); i++) {
    if (ans[i].r - ans[i].l + 1 >= x) return ans[i].val;
    x -= (ans[i].r - ans[i].l + 1);
  }
  return -1;
}
long long op4(long long l, long long r, long long x, long long y) {
  long long res = 0;
  cut(l);
  if (r + 1 <= n) cut(r + 1);
  node temp(l, inf, inf);
  set<node>::iterator it = S.upper_bound(temp);
  set<node>::iterator tmp;
  it--;
  while (it != S.end()) {
    if (it->r > r) break;
    node t = *it;
    it++;
    res = (res + pow(t.val, x, y) * (t.r - t.l + 1) % y) % y;
  }
  return res;
}
int main() {
  int i, j, k;
  cin >> n >> m >> seed >> vmax;
  node t(0, n, inf);
  S.insert(t);
  t.l = n + 1;
  S.insert(t);
  for (i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
    op2((long long)i, (long long)i, a[i]);
  }
  for (i = 1; i <= m; i++) {
    long long op = (rnd() % 4LL) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    long long x, y;
    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;
    if (op == 1) op1(l, r, x);
    if (op == 2) op2(l, r, x);
    if (op == 3) printf("%lld\n", op3(l, r, x));
    if (op == 4) printf("%lld\n", op4(l, r, x, y));
  }
  return 0;
}