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;
typedef struct NODE {
  long long a, b, la, lb;
  double ans, lans;
} NODE;
NODE operator+(NODE &a, NODE &b) {
  NODE c;
  c.a = a.a;
  c.b = a.b;
  c.la = b.la;
  c.lb = b.lb;
  double r = b.ans / (1 - (1 - b.ans) * a.lans);
  c.ans = a.ans * b.ans * (1 / (1 - a.lans * (1 - b.ans)));
  c.lans = b.lans + (1 - b.lans) * a.lans * r;
  return c;
}
long long n, q;
pair<long long, long long> pr[100005];
NODE tr[4 * 100005];
void build(long long, long long, long long),
    upd(long long, long long, long long, long long);
NODE query(long long, long long, long long, long long, long long);
int main() {
  scanf("%lld %lld", &n, &q);
  for (int i = 1; i < n + 1; i++)
    scanf("%lld %lld", &pr[i].first, &pr[i].second);
  build(0, 1, n);
  long long t, x, a, b;
  for (int i = 0; i < q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> x >> a >> b;
      pr[x].first = a, pr[x].second = b;
      upd(0, 1, n, x);
    } else {
      cin >> a >> b;
      printf("%0.12lf\n", query(0, 1, n, a, b).ans);
    }
  }
}
void build(long long idx, long long l, long long r) {
  if (l == r) {
    tr[idx].a = pr[l].first;
    tr[idx].b = pr[l].second;
    tr[idx].la = pr[l].first;
    tr[idx].lb = pr[l].second;
    tr[idx].ans = tr[idx].lans = (pr[l].first / (pr[l].second + 0.0));
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * idx + 1, l, mid);
  build(2 * idx + 2, mid + 1, r);
  tr[idx] = (tr[2 * idx + 1] + tr[2 * idx + 2]);
}
void upd(long long idx, long long l, long long r, long long pos) {
  if (r < pos || pos < l) return;
  if (l == r) {
    tr[idx].a = pr[l].first;
    tr[idx].b = pr[l].second;
    tr[idx].la = pr[l].first;
    tr[idx].lb = pr[l].second;
    tr[idx].ans = tr[idx].lans = (pr[l].first / (pr[l].second + 0.0));
    return;
  }
  long long mid = (l + r) / 2;
  upd(2 * idx + 1, l, mid, pos);
  upd(2 * idx + 2, mid + 1, r, pos);
  tr[idx] = tr[2 * idx + 1] + tr[2 * idx + 2];
}
NODE query(long long idx, long long l, long long r, long long low,
           long long high) {
  if (low <= l && r <= high) return tr[idx];
  long long mid = (l + r) / 2;
  if (high <= mid)
    return query(2 * idx + 1, l, mid, low, high);
  else if (mid + 1 <= low)
    return query(2 * idx + 2, mid + 1, r, low, high);
  NODE a = query(2 * idx + 1, l, mid, low, high);
  NODE b = query(2 * idx + 2, mid + 1, r, low, high);
  return (a + b);
}