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
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
pair<long double, long double> st[4 * N];
int n, q;
template <typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) {
  pair<T, T> res;
  res.first = a.first + b.first * a.second;
  res.second = a.second * b.second;
  return res;
}
void update(int id, int l, int r, int pos, long double val) {
  if (l > pos || r < pos) return;
  if (l == r) {
    st[id] = make_pair(val, val);
    return;
  }
  int mid = (l + r) >> 1;
  update(id << 1, l, mid, pos, val);
  update(id << 1 | 1, mid + 1, r, pos, val);
  st[id] = st[id << 1] + st[id << 1 | 1];
}
pair<long double, long double> getsum(int id, int l, int r, int u, int v) {
  if (l > v || r < u) return make_pair(0, 1);
  if (l >= u && r <= v) return st[id];
  int mid = (l + r) >> 1;
  pair<long double, long double> lnode = getsum(id << 1, l, mid, u, v);
  pair<long double, long double> rnode = getsum(id << 1 | 1, mid + 1, r, u, v);
  return lnode + rnode;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    long double p = (long double)a / b;
    update(1, 1, n, i, (1.0L - p) / p);
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int pos, a, b;
      cin >> pos >> a >> b;
      long double p = (long double)a / b;
      update(1, 1, n, pos, (1.0L - p) / p);
    } else {
      int l, r;
      cin >> l >> r;
      long double res = 1 + getsum(1, 1, n, l, r).first;
      long double ans = (1.0L / res);
      if (ans != ans) ans = 0;
      cout << fixed << setprecision(9) << ans << "\n";
    }
  }
  return 0;
}
