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 SIZE = 100100;
double a[SIZE];
pair<double, double> tree[4 * SIZE];
int n, q, gl, gr;
pair<double, double> merge(pair<double, double> p1, pair<double, double> p2) {
  return {(p1.first * p2.first) / (1 - (1 - p2.first) * p1.second),
          p2.second + (1 - p2.second) * p1.second * p2.first /
                          (1 - (1 - p2.first) * p1.second)};
}
void build(int node, int ll, int lr) {
  if (ll > lr) return;
  if (ll == lr) {
    tree[node] = {a[ll], a[ll]};
    return;
  }
  build(node * 2, ll, (ll + lr) / 2);
  build(node * 2 + 1, 1 + (ll + lr) / 2, lr);
  tree[node] = merge({tree[node * 2].first, tree[node * 2].second},
                     {tree[node * 2 + 1].first, tree[node * 2 + 1].second});
}
void update(int node, int ll, int lr, double value) {
  if (ll > lr || ll > gr || lr < gl) return;
  if (ll == lr) {
    tree[node] = {value, value};
    return;
  }
  update(node * 2, ll, (ll + lr) / 2, value);
  update(node * 2 + 1, (ll + lr) / 2 + 1, lr, value);
  tree[node] = merge({tree[node * 2].first, tree[node * 2].second},
                     {tree[node * 2 + 1].first, tree[node * 2 + 1].second});
}
pair<double, double> query(int node, int ll, int lr) {
  if (ll > lr || ll > gr || lr < gl) return {1, 0};
  if (ll >= gl && lr <= gr) return tree[node];
  return merge(query(node * 2, ll, (ll + lr) / 2),
               query(node * 2 + 1, 1 + (ll + lr) / 2, lr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  cin >> n >> q;
  int x, y, type;
  for (int i = 0; i < n; i++) cin >> x >> y, a[i] = double(x) / y;
  build(1, 0, n - 1);
  while (q--) {
    cin >> type;
    if (type == 1) {
      cin >> gl;
      gl--;
      gr = gl;
      cin >> x >> y;
      update(1, 0, n - 1, double(x) / y);
    } else {
      cin >> gl >> gr;
      gl--, gr--;
      cout << query(1, 0, n - 1).first << '\n';
    }
  }
  return 0;
}