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 readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct node {
  double x, y;
} t[300000];
int n, q;
int a[100005], b[100005];
void update(int id) {
  t[id].x = t[id << 1].x * t[id << 1 | 1].x /
            (1 - (1 - t[id << 1 | 1].x) * t[id << 1].y);
  t[id].y = t[id << 1 | 1].y + (1 - t[id << 1 | 1].y) * t[id << 1].y *
                                   t[id << 1 | 1].x /
                                   (1 - (1 - t[id << 1 | 1].x) * t[id << 1].y);
}
void build(int id, int l, int r) {
  if (l == r) return (void)(t[id].x = t[id].y = 1.0 * a[l] / b[l]);
  int mid = (l + r) / 2;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  update(id);
}
void change(int id, int l, int r, int p, double c) {
  if (l == p && r == p) return (void)(t[id].x = t[id].y = c);
  int mid = (l + r) / 2;
  if (p <= mid)
    change(id << 1, l, mid, p, c);
  else
    change(id << 1 | 1, mid + 1, r, p, c);
  update(id);
}
pair<double, double> query(int id, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return make_pair(t[id].x, t[id].y);
  int mid = (l + r) / 2;
  if (qr <= mid)
    return query(id << 1, l, mid, ql, qr);
  else if (ql > mid)
    return query(id << 1 | 1, mid + 1, r, ql, qr);
  pair<double, double> lson = query(id << 1, l, mid, ql, mid),
                       rson = query(id << 1 | 1, mid + 1, r, mid + 1, qr), ans;
  ans.first = lson.first * rson.first / (1 - (1 - rson.first) * lson.second);
  ans.second = rson.second + (1 - rson.second) * lson.second * rson.first /
                                 (1 - (1 - rson.first) * lson.second);
  return ans;
}
int main() {
  n = readint();
  q = readint();
  for (int i = 1; i <= n; i++) a[i] = readint(), b[i] = readint();
  build(1, 1, n);
  int opt, x, y;
  while (q--) {
    opt = readint();
    x = readint();
    y = readint();
    if (opt == 1)
      change(1, 1, n, x, 1.0 * y / readint());
    else
      printf("%.10lf\n", query(1, 1, n, x, y).first);
  }
  return 0;
}