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 N = 100100;
pair<double, double> tree[4 * N];
double p[N];
int n, q;
pair<double, double> mer(pair<double, double> a, pair<double, double> b) {
  pair<double, double> ans;
  ans.first = a.first * b.first / (1 - a.second + a.second * b.first);
  ans.second = b.second + (1 - b.second) * a.second * b.first /
                              (1 - a.second + a.second * b.first);
  return ans;
}
void buildTree(int s, int l, int r) {
  if (l == r) {
    tree[s].first = tree[s].second = p[l];
    return;
  }
  int mid = (l + r) >> 1;
  buildTree(2 * s, l, mid);
  buildTree(2 * s + 1, mid + 1, r);
  tree[s] = mer(tree[2 * s], tree[2 * s + 1]);
}
void update(int s, int l, int r, int u, double w) {
  if (l > u || r < u) return;
  if (l == r) {
    tree[s].first = tree[s].second = w;
    return;
  }
  int mid = (l + r) >> 1;
  update(2 * s, l, mid, u, w);
  update(2 * s + 1, mid + 1, r, u, w);
  tree[s] = mer(tree[2 * s], tree[2 * s + 1]);
}
pair<double, double> get(int s, int l, int r, int u, int v) {
  if (r < u || l > v) return make_pair(1, 0);
  if (u <= l && r <= v) return tree[s];
  int mid = (l + r) >> 1;
  pair<double, double> t1 = get(2 * s, l, mid, u, v);
  pair<double, double> t2 = get(2 * s + 1, mid + 1, r, u, v);
  return mer(t1, t2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    p[i] = 1.0 * a / b;
  }
  buildTree(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int type, l, r, v;
    cin >> type;
    if (type == 1) {
      cin >> v >> l >> r;
      update(1, 1, n, v, 1.0 * l / r);
    } else {
      cin >> l >> r;
      pair<double, double> ans = get(1, 1, n, l, r);
      cout << fixed << setprecision(8) << ans.first << endl;
    }
  }
  return 0;
}