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;
int n, q;
double tree[400005];
double tree2[400005];
double p[400005];
double m(double x, double y, double x2) {
  if (x < -1) return y;
  if (y < -1) return x;
  return x * y * (1 / (1 - ((1 - x2) * (1 - y))));
}
void build(int i, int l, int r) {
  if (l == r) {
    tree[i] = p[l];
    tree2[i] = 1 - p[l];
    return;
  }
  build(i * 2 + 1, l, (l + r) / 2);
  build(i * 2 + 2, (l + r) / 2 + 1, r);
  tree[i] = m(tree[i * 2 + 1], tree[i * 2 + 2], tree2[i * 2 + 1]);
  tree2[i] = m(tree2[i * 2 + 2], tree2[i * 2 + 1], tree[i * 2 + 2]);
}
void upd(int i, int l, int r, int x, double val) {
  if (l == r && l == x) {
    tree[i] = val;
    tree2[i] = 1 - val;
    return;
  }
  if (r < x || l > x) return;
  upd(i * 2 + 1, l, (l + r) / 2, x, val);
  upd(i * 2 + 2, (l + r) / 2 + 1, r, x, val);
  tree[i] = m(tree[i * 2 + 1], tree[i * 2 + 2], tree2[i * 2 + 1]);
  tree2[i] = m(tree2[i * 2 + 2], tree2[i * 2 + 1], tree[i * 2 + 2]);
}
pair<double, double> query(int i, int l, int r, int x, int y) {
  if (l > y || r < x) return make_pair(-2.0, -2.0);
  if (l >= x && r <= y) return make_pair(tree[i], tree2[i]);
  pair<double, double> L = query(i * 2 + 1, l, (l + r) / 2, x, y);
  pair<double, double> R = query(i * 2 + 2, (l + r) / 2 + 1, r, x, y);
  pair<double, double> res;
  res.first = m(L.first, R.first, L.second);
  res.second = m(R.second, L.second, R.first);
  return res;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    p[i] = (1.0 * a) / b;
  }
  build(0, 0, n - 1);
  while (q--) {
    int com;
    scanf("%d", &com);
    if (com == 1) {
      int i, a, b;
      scanf("%d %d %d", &i, &a, &b);
      i--;
      p[i] = (1.0 * a) / b;
      upd(0, 0, n - 1, i, p[i]);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      r--;
      printf("%.7lf\n", query(0, 0, n - 1, l, r).first);
    }
  }
}