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 int MAX = 1e5 + 5;
const long double EPS = 1e-12;
long double arr[MAX], u[MAX], p[MAX];
pair<long double, long double> st[4 * MAX];
int n;
inline pair<long double, long double> merge(pair<long double, long double> f,
                                            pair<long double, long double> s) {
  pair<long double, long double> ans;
  ans.first = f.first * s.first;
  ans.second = f.second + s.second * f.first;
  return ans;
}
void build(int id, int l, int r) {
  if (l == r) {
    st[id] = make_pair(u[l], u[l]);
    return;
  }
  build(id * 2, l, (l + r) / 2);
  build(id * 2 + 1, (l + r) / 2 + 1, r);
  st[id] = merge(st[id * 2], st[id * 2 + 1]);
}
void update(int id, int l, int r, int idx) {
  if (l == r && l == idx) {
    st[id] = make_pair(u[l], u[l]);
    return;
  }
  if (idx > r || l > idx) {
    return;
  }
  int mid = (l + r) / 2;
  update(id * 2, l, mid, idx);
  update(id * 2 + 1, mid + 1, r, idx);
  st[id] = merge(st[id * 2], st[id * 2 + 1]);
}
pair<long double, long double> query(int id, int l, int r, int i, int j) {
  if (l >= i && j >= r) {
    return st[id];
  }
  if (i > r || l > j) {
    return make_pair(1.0, 0.0);
  }
  int mid = (l + r) / 2;
  auto q1 = query(id * 2, l, mid, i, j);
  auto q2 = query(id * 2 + 1, mid + 1, r, i, j);
  auto ret = merge(q1, q2);
  return ret;
}
int main() {
  int q;
  scanf("%d %d", &n, &q);
  for (int i = int(1); i < int(n + 1); i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    p[i] = (long double)a / (long double)b;
    u[i] = (1 - p[i]) / p[i];
  }
  build(1, 1, n);
  for (int i = int(1); i < int(q + 1); i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int id, a, b;
      scanf("%d %d %d", &id, &a, &b);
      p[id] = (long double)a / (long double)b;
      u[id] = (1 - p[id]) / p[id];
      update(1, 1, n, id);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      long double cur = query(1, 1, n, l, r).second;
      long double ans = 0.0;
      if (cur < 1e30) {
        ans = 1.0 / (1.0 + cur);
      }
      printf("%.10Lf\n", ans);
    }
  }
  return 0;
}