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 seed, MOD = 1e9 + 7;
long long rnd() {
  long long ret = seed;
  seed = seed * 7 + 13;
  seed %= MOD;
  return ret;
}
long long a[121212];
struct QR {
  long long op, l, r, x, y;
} qr[121212];
long long pow(long long a, long long b, long long mod) {
  if (b == 1) return a % mod;
  long long x = pow(a, b / 2, mod);
  return ((x * x % mod) * (b % 2 == 1 ? a % mod : 1)) % mod;
}
set<pair<pair<long long, long long>, long long>> S;
int main() {
  long long n, m, vmax;
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  long long i, j;
  for (i = 1; i <= n; i++) a[i] = rnd() % vmax + 1;
  for (i = 1; i <= m; i++) {
    long long op = rnd() % 4 + 1;
    long long l = rnd() % n + 1;
    long long r = rnd() % n + 1;
    if (l > r) swap(l, r);
    long long x, y = 0;
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    qr[i] = {op, l, r, x, y};
  }
  for (i = 1; i <= n; i++) S.insert({{i, i}, a[i]});
  for (i = 1; i <= m; i++) {
    long long op = qr[i].op, l = qr[i].l, r = qr[i].r, x = qr[i].x, y = qr[i].y;
    auto w = S.lower_bound({{l, -MOD}, -MOD});
    if (op == 1 || op == 2) {
      while (w != S.end()) {
        long long e = w->first.first, s = w->first.second, v = w->second;
        if (r < s) break;
        auto nxt = w;
        nxt++;
        S.erase(w);
        if (s < l) {
          S.insert({{l - 1, s}, v});
          s = l;
        }
        if (r < e) {
          S.insert({{e, r + 1}, v});
          e = r;
        }
        if (op == 1) S.insert({{e, s}, v + x});
        w = nxt;
      }
      if (op == 2) S.insert({{r, l}, x});
    }
    if (op == 3 || op == 4) {
      vector<pair<long long, pair<long long, long long>>> L;
      auto w = S.lower_bound({{l, -MOD}, -MOD});
      while (w != S.end()) {
        if (r < w->first.second) break;
        L.push_back({w->second, {w->first.second, w->first.first}});
        w++;
      }
      long long cnt = 0;
      if (op == 3) {
        sort(L.begin(), L.end());
        for (auto v : L) {
          long long s = max(l, v.second.first), e = min(r, v.second.second),
                    g = v.first;
          cnt += e - s + 1;
          if (cnt >= x) {
            printf("%lld\n", g);
            break;
          }
        }
      }
      if (op == 4) {
        long long sum = 0;
        for (auto v : L) {
          long long s = max(l, v.second.first), e = min(r, v.second.second),
                    g = v.first % y;
          sum = (sum + (pow(g, x, y) * (e - s + 1)) % y) % y;
        }
        printf("%lld\n", sum);
      }
    }
  }
  return 0;
}