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;
void Freopen() {
  freopen(
      "title"
      ".in",
      "r", stdin);
  freopen(
      "title"
      ".out",
      "w", stdout);
}
int read() {
  int g = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    g = g * 10 + ch - '0';
    ch = getchar();
  }
  return g * f;
}
const int N = 1e5 + 5;
int n, q;
double p[N], t[N];
struct node {
  double d, al;
};
node operator+(node a, node b) {
  node now;
  now.al = a.al * b.al;
  now.d = a.d + b.d * a.al;
  return now;
}
struct Seg {
  node tr[N << 2];
  void up(int v) { tr[v] = tr[v << 1] + tr[v << 1 | 1]; }
  void build(int v, int l, int r) {
    if (l == r) return tr[v] = node{t[l], t[l]}, void();
    int mid = l + r >> 1;
    build(v << 1, l, mid), build(v << 1 | 1, mid + 1, r);
    up(v);
  }
  void insert(int v, int l, int r, int q, int w) {
    if (l == r) return tr[v] = node{t[l], t[l]}, void();
    int mid = l + r >> 1;
    q <= mid ? insert(v << 1, l, mid, q, w)
             : insert(v << 1 | 1, mid + 1, r, q, w);
    up(v);
  }
  node query(int v, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return tr[v];
    int mid = l + r >> 1;
    if (qr <= mid) return query(v << 1, l, mid, ql, qr);
    if (ql > mid) return query(v << 1 | 1, mid + 1, r, ql, qr);
    return query(v << 1, l, mid, ql, qr) +
           query(v << 1 | 1, mid + 1, r, ql, qr);
  }
} seg;
signed main() {
  n = read(), q = read();
  for (int i = (1); i <= (n); i++) p[i] = 1. * read() / read();
  for (int i = (1); i <= (n); i++) t[i] = (1 - p[i]) / p[i];
  seg.build(1, 1, n);
  for (int ca = (1); ca <= (q); ca++) {
    int op = read();
    if (op == 1) {
      int x = read(), a = read(), b = read();
      p[x] = 1. * a / b;
      t[x] = (1 - p[x]) / p[x];
      seg.insert(1, 1, n, x, t[x]);
    } else {
      int l = read(), r = read();
      cout << fixed << setprecision(10) << 1 / (seg.query(1, 1, n, l, r).d + 1)
           << '\n';
    }
  }
  return signed();
}