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 n, q;
const int S = 1 << 17;
pair<long double, long double> tree[2 * S];
pair<long double, long double> merge(pair<long double, long double> a,
                                     pair<long double, long double> b) {
  return make_pair(
      a.first * b.first / (1 - (1.0 - a.second) * (1 - b.first)),
      a.second * b.second / (1 - (1.0 - a.second) * (1 - b.first)));
}
pair<long double, long double> interRec(int p, int q, int pp, int qq,
                                        int node) {
  if (q < pp || qq < p) {
    return make_pair(1.0, 1.0);
  }
  if (p <= pp && qq <= q) {
    return tree[node];
  }
  int ss = (pp + qq) / 2;
  return merge(interRec(p, q, pp, ss, 2 * node),
               interRec(p, q, ss + 1, qq, 2 * node + 1));
}
long double inter(int p, int q) {
  auto tmp = interRec(p, q, 0, S - 1, 1);
  return tmp.first;
}
void sett(int p, long double prob) {
  p += S;
  tree[p] = make_pair(prob, 1 - prob);
  p /= 2;
  while (p) {
    tree[p] = merge(tree[2 * p], tree[2 * p + 1]);
    p /= 2;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout << setprecision(20) << fixed;
  cin >> n >> q;
  int a, b;
  long double tmp;
  for (int i = 0; i < n; ++i) {
    cin >> a >> b;
    tmp = ((long double)a) / b;
    tree[i + S] = make_pair(tmp, 1 - tmp);
  }
  for (int i = n; i < S; ++i) {
    tree[i + S] = make_pair(1.0, 1.0);
  }
  for (int i = S - 1; i > 0; --i) {
    tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
  }
  for (int i = 0; i < q; ++i) {
    int type;
    cin >> type;
    if (type == 1) {
      int pos;
      cin >> pos >> a >> b;
      tmp = ((long double)a) / b;
      --pos;
      sett(pos, tmp);
    } else {
      int l, r;
      cin >> l >> r;
      --l;
      --r;
      cout << inter(l, r) << endl;
    }
  }
}