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;
inline long long rd() {
  long long x = 0;
  int ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void rt(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10)
    rt(x / 10), putchar(x % 10 + '0');
  else
    putchar(x + '0');
}
const int maxn = 100005;
pair<double, double> a[maxn << 2];
double p[maxn];
void update(int i) {
  double gl = (1 - (1 - a[(i << 1)].second) * (1 - a[(i << 1 | 1)].first));
  a[i].first = a[(i << 1)].first * a[(i << 1 | 1)].first / gl;
  a[i].second = a[(i << 1 | 1)].second * a[(i << 1)].second / gl;
}
void build(int i, int l, int r) {
  if (l == r) return a[i].first = p[l], a[i].second = 1 - p[l], void();
  int mid = l + r >> 1;
  build((i << 1), l, mid), build((i << 1 | 1), mid + 1, r);
  update(i);
}
void modify(int i, int l, int r, int ps, double v) {
  if (l == r) return a[i].first = v, a[i].second = 1 - v, void();
  int mid = l + r >> 1;
  if (ps <= mid)
    modify((i << 1), l, mid, ps, v);
  else
    modify((i << 1 | 1), mid + 1, r, ps, v);
  update(i);
}
pair<double, double> query(int i, int l, int r, int fr, int bh) {
  if (l == fr && r == bh) return a[i];
  int mid = l + r >> 1;
  if (bh <= mid)
    return query((i << 1), l, mid, fr, bh);
  else if (fr > mid)
    return query((i << 1 | 1), mid + 1, r, fr, bh);
  else {
    pair<double, double> L = query((i << 1), l, mid, fr, mid),
                         R = query((i << 1 | 1), mid + 1, r, mid + 1, bh);
    double gl = (1 - (1 - L.second) * (1 - R.first));
    return make_pair(L.first * R.first / gl, L.second * R.second / gl);
  }
}
int n, q;
int main() {
  n = rd(), q = rd();
  for (int i = 1; i <= (int)n; i++) {
    int fz = rd(), fm = rd();
    p[i] = 1.0 * fz / fm;
  }
  build(1, 1, n);
  for (; q--;) {
    int opt = rd();
    if (opt == 2) {
      int l = rd(), r = rd();
      printf("%.10lf\n", query(1, 1, n, l, r).first);
    } else {
      int ps = rd(), fz = rd(), fm = rd();
      modify(1, 1, n, ps, 1.0 * fz / fm);
    }
  }
}