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;
double a[100005], b[100005];
double t[400005][2];
void build(long long v, long long l, long long r) {
  if (l == r) {
    t[v][0] = a[l];
    t[v][0] /= b[l];
    t[v][1] = t[v][0];
    return;
  }
  long long m = (l + r) / 2;
  build(2 * v, l, m);
  build(2 * v + 1, m + 1, r);
  double l1 = t[2 * v][0];
  double l2 = t[2 * v + 1][0];
  double r1 = t[2 * v][1];
  double r2 = t[2 * v + 1][1];
  t[v][0] = l1 * l2;
  t[v][0] /= (1.0 - (1.0 - l2) * r1);
  t[v][1] = r2 + (r1 * l2 * (1.0 - r2)) / (1.0 - (1.0 - l2) * r1);
  return;
}
pair<double, double> get(long long v, long long l, long long r, long long ll,
                         long long rr) {
  if (l == ll && r == rr) {
    return {t[v][0], t[v][1]};
  }
  long long m = (l + r) / 2;
  if (rr <= m)
    return get(2 * v, l, m, ll, rr);
  else if (ll > m)
    return get(2 * v + 1, m + 1, r, ll, rr);
  else {
    pair<double, double> p1 = get(2 * v, l, m, ll, m);
    pair<double, double> p2 = get(2 * v + 1, m + 1, r, m + 1, rr);
    double l1 = p1.first;
    double l2 = p2.first;
    double r1 = p1.second;
    double r2 = p2.second;
    p1.first = l1 * l2;
    p1.first /= (1.0 - (1.0 - l2) * r1);
    p1.second = r2 + (r1 * l2 * (1.0 - r2)) / (1.0 - (1.0 - l2) * r1);
    return p1;
  }
}
void update(long long v, long long l, long long r, long long p, double x) {
  if (l == r) {
    t[v][0] = x;
    t[v][1] = x;
    return;
  }
  long long m = (l + r) / 2;
  if (p <= m)
    update(2 * v, l, m, p, x);
  else
    update(2 * v + 1, m + 1, r, p, x);
  double l1 = t[2 * v][0];
  double l2 = t[2 * v + 1][0];
  double r1 = t[2 * v][1];
  double r2 = t[2 * v + 1][1];
  t[v][0] = l1 * l2;
  t[v][0] /= (1.0 - (1.0 - l2) * r1);
  t[v][1] = r2 + (r1 * l2 * (1.0 - r2)) / (1.0 - (1.0 - l2) * r1);
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) cin >> a[i] >> b[i];
  build(1, 0, n - 1);
  long long x, y, z;
  for (long long i = 0; i < q; i++) {
    cin >> x;
    if (x == 1) {
      cin >> x >> y >> z;
      update(1, 0, n - 1, x - 1, (1.0 * y) / z);
    } else {
      cin >> x >> y;
      cout << setprecision(10) << get(1, 0, n - 1, x - 1, y - 1).first << "\n";
    }
  }
  return 0;
}