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;
pair<double, double> a[100005];
double st1[400004], st2[400005];
void refresh(int n) {
  double l1 = st1[2 * n], l2 = st1[2 * n + 1], r1 = st2[2 * n],
         r2 = st2[2 * n + 1];
  st1[n] = (l1 * l2) / (1 - ((1 - l2) * r1));
  st2[n] = r2 + ((r1 * l2 * (1 - r2)) / (1 - ((1 - l2) * r1)));
}
void build(int n, int l, int r) {
  if (l > r) return;
  if (l == r) {
    st1[n] = a[l].first / a[l].second;
    st2[n] = a[l].first / a[l].second;
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * n, l, mid);
  build(2 * n + 1, mid + 1, r);
  refresh(n);
}
void update(int n, int l, int r, int x, double y) {
  if (l > r || x < l || x > r) return;
  if (l == r) {
    st1[n] = y;
    st2[n] = y;
    return;
  }
  int mid = (l + r) >> 1;
  update(2 * n, l, mid, x, y);
  update(2 * n + 1, mid + 1, r, x, y);
  refresh(n);
}
pair<double, double> query(int n, int l, int r, int x, int y) {
  if (l > r || l > y || r < x)
    return pair<double, double>(make_pair(-1.0, -1.0));
  if (l >= x && r <= y) return pair<double, double>(make_pair(st1[n], st2[n]));
  int mid = (l + r) >> 1;
  pair<double, double> p1 = query(2 * n, l, mid, x, y);
  pair<double, double> p2 = query(2 * n + 1, mid + 1, r, x, y);
  if (p1 == pair<double, double>(make_pair(-1.0, -1.0))) return p2;
  if (p2 == pair<double, double>(make_pair(-1.0, -1.0))) return p1;
  pair<double, double> p;
  double l1 = p1.first, l2 = p2.first, r1 = p1.second, r2 = p2.second;
  p.first = (l1 * l2) / (1 - ((1 - l2) * r1));
  p.second = r2 + ((r1 * l2 * (1 - r2)) / (1 - ((1 - l2) * r1)));
  return p;
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
  build(1, 0, n - 1);
  int x, y, z, tmp;
  for (int i = 0; i < q; i++) {
    cin >> tmp;
    if (tmp == 1) {
      cin >> x >> y >> z;
      double tmp1 = (double)y / (double)z;
      x--;
      update(1, 0, n - 1, x, tmp1);
    } else {
      cin >> x >> y;
      x--;
      y--;
      cout << setprecision(8) << query(1, 0, n - 1, x, y).first << "\n";
    }
  }
}