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;
pair<double, double> stree[500005];
double aa[100005], bb[100005];
int n;
void build(int pos, int l, int r) {
  if (l == r) {
    double k = 0.0;
    k = ((double)aa[l] / (double)bb[l]);
    stree[pos].first = k;
    stree[pos].second = k;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * pos + 1, l, mid);
  build(2 * pos + 2, mid + 1, r);
  stree[pos].first = (stree[2 * pos + 1].first * stree[2 * pos + 2].first) /
                     (1.00 - stree[2 * pos + 1].second +
                      stree[2 * pos + 1].second * stree[2 * pos + 2].first);
  stree[pos].second =
      (stree[2 * pos + 2].second +
       (((1.00 - stree[2 * pos + 2].second) * stree[2 * pos + 1].second *
         stree[2 * pos + 2].first) /
        (1.00 - stree[2 * pos + 1].second +
         stree[2 * pos + 1].second * stree[2 * pos + 2].first)));
}
void update(int pos, int l, int r, int xx, double a, double b) {
  if (l == r) {
    double k = 0.0;
    k = ((double)a / (double)b);
    stree[pos].first = k;
    stree[pos].second = k;
    return;
  }
  int mid = (l + r) / 2;
  if (xx < mid + 1)
    update(2 * pos + 1, l, mid, xx, a, b);
  else
    update(2 * pos + 2, mid + 1, r, xx, a, b);
  stree[pos].first = (stree[2 * pos + 1].first * stree[2 * pos + 2].first) /
                     (1.00 - stree[2 * pos + 1].second +
                      stree[2 * pos + 1].second * stree[2 * pos + 2].first);
  stree[pos].second =
      (stree[2 * pos + 2].second +
       (((1.00 - stree[2 * pos + 2].second) * stree[2 * pos + 1].second *
         stree[2 * pos + 2].first) /
        (1.00 - stree[2 * pos + 1].second +
         stree[2 * pos + 1].second * stree[2 * pos + 2].first)));
}
pair<double, double> query(int pos, int l, int r, int first, int second) {
  if (first <= l && r <= second) {
    if (l == 0 && r == n - 1) printf("%.10lf\n", stree[pos].first);
    return stree[pos];
  }
  if (l > second || r < first) return make_pair(-1, -1);
  int mid = (l + r) / 2;
  pair<double, double> a = query(2 * pos + 1, l, mid, first, second);
  pair<double, double> b = query(2 * pos + 2, mid + 1, r, first, second);
  pair<double, double> ans;
  if (a.first == -1) {
    if (l == 0 && r == n - 1) printf("%.10lf\n", b.first);
    return b;
  } else if (b.first == -1) {
    if (l == 0 && r == n - 1) printf("%.10lf\n", a.first);
    return a;
  }
  ans.first = (a.first * b.first) / (1.00 - a.second + a.second * b.first);
  ans.second = (b.second + (((1.00 - b.second) * a.second * b.first) /
                            (1.00 - a.second + a.second * b.first)));
  if (l == 0 && r == n - 1) printf("%.10lf\n", ans.first);
  return ans;
}
int main() {
  int q;
  cin >> n >> q;
  int i;
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> aa[i] >> bb[i];
  build(0, 0, n - 1);
  while (q--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int xx;
      double a, b;
      cin >> xx >> a >> b;
      xx--;
      update(0, 0, n - 1, xx, a, b);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      pair<int, int> ans;
      ans = query(0, 0, n - 1, l, r);
    }
  }
  return 0;
}