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 N = 1e5 + 500;
const pair<double, double> inf = make_pair(-1, -1);
pair<double, double> T[N << 2];
double f[N];
int n, m;
inline int read() {
  int n = 0;
  char c;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
  return n;
}
pair<double, double> merg(pair<double, double> t1, pair<double, double> t2) {
  double l1 = t1.first, r1 = t1.second, l2 = t2.first, r2 = t2.second;
  return make_pair(l1 * l2 / (1 + (l2 - 1) * r1),
                   r2 + l2 * (1 - r2) * r1 / (1 - (1 - l2) * r1));
}
void up(int x) { T[x] = merg(T[x << 1], T[x << 1 | 1]); }
void build(int x, int l, int r) {
  if (l == r) {
    T[x] = make_pair(f[l], f[l]);
    return;
  }
  int mi = (l + r) >> 1;
  build(x << 1, l, mi);
  build(x << 1 | 1, mi + 1, r);
  up(x);
}
void modify(int x, int l, int r, int pos) {
  if (l == r) {
    T[x] = make_pair(f[l], f[l]);
    return;
  }
  int mi = (l + r) >> 1;
  if (pos <= mi)
    modify(x << 1, l, mi, pos);
  else
    modify(x << 1 | 1, mi + 1, r, pos);
  up(x);
}
pair<double, double> qry(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) return T[x];
  int mi = (l + r) >> 1;
  pair<double, double> t1 = inf, t2 = inf;
  if (L <= mi) t1 = qry(x << 1, l, mi, L, R);
  if (mi < R) t2 = qry(x << 1 | 1, mi + 1, r, L, R);
  return t1 == inf ? t2 : (t2 == inf ? t1 : merg(t1, t2));
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    f[i] = x * 1.0 / y;
  }
  build(1, 1, n);
  while (m--) {
    int kd, x, y, z;
    kd = read(), x = read(), y = read();
    if (kd == 2) {
      pair<double, double> ans = qry(1, 1, n, x, y);
      printf("%.10lf\n", ans.first);
    } else {
      z = read();
      f[x] = y * 1.0 / z;
      modify(1, 1, n, x);
    }
  }
  return 0;
}