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 mx = 100500;
int a[mx];
int b[mx];
pair<double, double> p[mx * 4 + 5];
int root = 1;
pair<double, double> recalc(pair<double, double> lhs,
                            pair<double, double> rhs) {
  double p1 = lhs.first;
  double q1 = lhs.second;
  double p2 = rhs.first;
  double q2 = rhs.second;
  double p = p1 * p2 / (1 - (1 - q1) * (1 - p2));
  swap(p1, q1);
  swap(p2, q2);
  swap(p1, p2);
  swap(q1, q2);
  double q = p1 * p2 / (1 - (1 - q1) * (1 - p2));
  return {p, q};
}
void recalc(int u) { p[u] = recalc(p[2 * u], p[2 * u + 1]); }
void build(int u, int l, int r) {
  if (l == r) {
    p[u].first = double(a[l]) / b[l];
    p[u].second = 1 - p[u].first;
    return;
  }
  int d = (l + r) >> 1;
  build(2 * u, l, d);
  build(2 * u + 1, d + 1, r);
  recalc(u);
}
void update(int u, int l, int r, int pos, double val) {
  if (pos < l || pos > r) return;
  if (l == r) {
    p[u].first = val;
    p[u].second = 1 - p[u].first;
    return;
  }
  int d = (l + r) >> 1;
  update(2 * u, l, d, pos, val);
  update(2 * u + 1, d + 1, r, pos, val);
  recalc(u);
}
bool intersect(int l, int r, int x, int y) {
  if (y < l) return false;
  if (r < x) return false;
  return true;
}
pair<double, double> get_value(int u, int l, int r, int x, int y) {
  if (l >= x && r <= y) {
    return p[u];
  }
  int d = (l + r) >> 1;
  if (!intersect(l, d, x, y)) return get_value(2 * u + 1, d + 1, r, x, y);
  if (!intersect(d + 1, r, x, y)) return get_value(2 * u, l, d, x, y);
  auto p1 = get_value(2 * u, l, d, x, y);
  auto p2 = get_value(2 * u + 1, d + 1, r, x, y);
  return recalc(p1, p2);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d%d", a + i, b + i);
  build(root, 1, n);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int a, b, pos;
      scanf("%d%d%d", &pos, &a, &b);
      update(root, 1, n, pos, double(a) / b);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.10f\n", get_value(root, 1, n, l, r).first);
    }
  }
  return 0;
}