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 long long mod = 10007;
const long long inf = 1e16;
const long long ninf = LLONG_MIN;
const long double eps = 1e-10;
const long long N = 1000005;
const long long LOGN = 19;
const long double PI = 3.14159265358979323846;
long long mul(long long a, long long b) { return (a * b) % mod; }
long long add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
  return a;
}
long long n, q;
long double p[N];
pair<long double, long double> st[N];
bool is(long double x) { return (abs(x) <= eps); }
pair<long double, long double> merge(pair<long double, long double> l,
                                     pair<long double, long double> r) {
  long double l1 = l.first, r1 = l.second, l2 = r.first, r2 = r.second;
  if (is(l1 + 1) && is(r1 + 1)) return r;
  if (is(l2 + 1) && is(r2 + 1)) return l;
  long double rec = (l2 - 1) * r1 + 1;
  if (is(rec)) return {0, 0};
  long double x = (l1 * l2) / (rec);
  long double y = r2 + (r1 * (1 - r2) * (l2)) / rec;
  return {x, y};
}
void update(long long id, long long s, long long e, long long x,
            long double v) {
  if (s == e) {
    st[id] = {v, v};
    p[s] = v;
    return;
  }
  long long m = (s + e) >> 1;
  if (x <= m) {
    update(id + id, s, m, x, v);
  } else {
    update(id + id + 1, m + 1, e, x, v);
  }
  st[id] = merge(st[id + id], st[id + id + 1]);
}
pair<long double, long double> query(long long id, long long s, long long e,
                                     long long l, long long r) {
  if (s > e || s > r || e < l) {
    return {-1, -1};
  }
  if (l <= s && e <= r) {
    return st[id];
  }
  long long m = (s + e) >> 1;
  return merge(query(id + id, s, m, l, r), query(id + id + 1, m + 1, e, l, r));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> q;
  cout << fixed << setprecision(11);
  for (int i = 0; i < n; i++) {
    long double v, a, b;
    cin >> a >> b;
    v = a / b;
    update(1, 0, n - 1, i, v);
  }
  for (int i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long double v, a, b, in;
      cin >> in >> a >> b;
      in--;
      v = a / b;
      update(1, 0, n - 1, in, v);
    } else {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << query(1, 0, n - 1, l, r).first << "\n";
    }
  }
  return 0;
}