Problem: There are n casinos lined in a row. If Memory plays at casino i, he has probability pi to win and move to the casino on the right (i + 1) or exit the row (if i = n), and a probability 1 - pi to lose and move to the casino on the left (i - 1) or also exit the row (if i = 1). 

We say that Memory dominates on the interval i... j if he completes a walk such that,

  * He starts on casino i. 
  * He never looses in casino i. 
  * He finishes his walk by winning in casino j. 



Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process.

Now Memory has some requests, in one of the following forms:

  * 1 i a b: Set <image>. 
  * 2 l r: Print the probability that Memory will dominate on the interval l... r, i.e. compute the probability that Memory will first leave the segment l... r after winning at casino r, if she starts in casino l. 



It is guaranteed that at any moment of time p is a non-decreasing sequence, i.e. pi ≤ pi + 1 for all i from 1 to n - 1.

Please help Memory by answering all his requests!

Input

The first line of the input contains two integers n and q(1 ≤ n, q ≤ 100 000), — number of casinos and number of requests respectively.

The next n lines each contain integers ai and bi (1 ≤ ai < bi ≤ 109) — <image> is the probability pi of winning in casino i.

The next q lines each contain queries of one of the types specified above (1 ≤ a < b ≤ 109, 1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ n).

It's guaranteed that there will be at least one query of type 2, i.e. the output will be non-empty. Additionally, it is guaranteed that p forms a non-decreasing sequence at all times.

Output

Print a real number for every request of type 2 — the probability that boy will "dominate" on that interval. Your answer will be considered correct if its absolute error does not exceed 10 - 4.

Namely: let's assume that one of your answers is a, and the corresponding answer of the jury is b. The checker program will consider your answer correct if |a - b| ≤ 10 - 4.

Example

Input

3 13
1 3
1 2
2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3
1 2 2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3


Output

0.3333333333
0.2000000000
0.1666666667
0.5000000000
0.4000000000
0.6666666667
0.3333333333
0.2500000000
0.2222222222
0.6666666667
0.5714285714
0.6666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1 << 17;
struct node {
  long double a, b;
  node(long double A = 1., long double B = 0.) : a(A), b(B) {}
};
vector<node> v(maxn * 2);
vector<long double> u(maxn);
vector<int> l(maxn * 2, 0), r(maxn * 2, 0);
node merge(const node &l, const node &r) {
  return node(l.a * r.a, l.b + r.b * l.a);
}
void upd(int i, int vr = 0) {
  if (l[vr] > i || r[vr] < i) return;
  if (l[vr] == i && r[vr] == i) {
    v[vr] = {u[i], u[i]};
    return;
  }
  upd(i, vr * 2 + 1), upd(i, vr * 2 + 2);
  v[vr] = merge(v[vr * 2 + 1], v[vr * 2 + 2]);
}
node query(int li, int ri, int vr = 0) {
  if (ri < l[vr] || r[vr] < li) return {1., 0.};
  if (li <= l[vr] && r[vr] <= ri) return v[vr];
  node nl = query(li, ri, vr * 2 + 1), nr = query(li, ri, vr * 2 + 2);
  return merge(nl, nr);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(15);
  for (int i = maxn - 1; i < maxn * 2; i++) l[i] = r[i] = i - (maxn - 1);
  for (int i = maxn - 2; i >= 0; i--) l[i] = l[i * 2 + 1], r[i] = r[i * 2 + 2];
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    long double a, b;
    cin >> a >> b;
    u[i] = (1. - a / b) / (a / b);
    upd(i);
  }
  while (q--) {
    int typ;
    cin >> typ;
    if (typ == 1) {
      int i;
      long double a, b;
      cin >> i >> a >> b;
      u[--i] = (1. - a / b) / (a / b);
      upd(i);
    } else {
      int l, r;
      cin >> l >> r;
      long double b = query(--l, --r).b;
      if (b < 1e30)
        cout << 1. / (1. + b) << "\n";
      else
        cout << "0\n";
    }
  }
  return 0;
}