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;
struct node {
  long double pL, pR;
};
node combine(node fi, node se) {
  if (fi.pL == -1) return se;
  if (se.pL == -1) return fi;
  node res;
  res.pL = fi.pL * se.pL / (1 + (se.pL - 1) * fi.pR);
  res.pR = se.pR + ((1 - se.pR) * fi.pR * se.pL) / (1 - fi.pR * (1 - se.pL));
  return res;
}
const int N = 100100;
long double a[N], b[N];
node t[N * 4];
int n;
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = {a[l] / b[l], a[l] / b[l]};
    return;
  }
  int mid = (l + r) >> 1;
  build(v + v, l, mid);
  build(v + v + 1, mid + 1, r);
  t[v] = combine(t[v + v], t[v + v + 1]);
}
void update(int v, int l, int r, int pos) {
  if (l == r) {
    t[v] = {a[l] / b[l], a[l] / b[l]};
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update(v + v, l, mid, pos);
  else
    update(v + v + 1, mid + 1, r, pos);
  t[v] = combine(t[v + v], t[v + v + 1]);
}
node get(int v, int l, int r, int tl, int tr) {
  if (l > r || tl > r || l > tr || tl > tr) return {-1, -1};
  if (tl <= l && r <= tr) {
    return t[v];
  }
  int mid = (l + r) >> 1;
  return combine(get(v + v, l, mid, tl, tr),
                 get(v + v + 1, mid + 1, r, tl, tr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> n >> tt;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
  }
  build(1, 1, n);
  cout.precision(10);
  cout << fixed;
  while (tt--) {
    int type;
    cin >> type;
    if (type == 2) {
      int l, r;
      cin >> l >> r;
      cout << get(1, 1, n, l, r).pL << "\n";
    } else {
      int pos;
      cin >> pos;
      cin >> a[pos] >> b[pos];
      update(1, 1, n, pos);
    }
  }
}