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 read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
inline int lowbit(int x) { return x & (-x); }
struct node {
  double ans, sum;
  node() { ans = 0, sum = 0; }
  node(double a, double b) { ans = a, sum = b; }
} t[400010];
int n, q;
double b[400010], a[400010];
inline node pushup(node a, node b) {
  return node(min(a.ans + b.ans * a.sum, (double)1e9),
              min(a.sum * b.sum, (double)1e9));
}
void build(int l, int r, int x) {
  if (l == r) {
    t[x].sum = t[x].ans = (b[l] - a[l]) / a[l];
    return;
  }
  build(l, ((l + r) >> 1), (x << 1));
  build(((l + r) >> 1) + 1, r, (x << 1 | 1));
  t[x] = pushup(t[(x << 1)], t[(x << 1 | 1)]);
}
void upd(int l, int r, int x, int pos) {
  if (l == r) {
    t[x].sum = t[x].ans = (b[l] - a[l]) / a[l];
    return;
  }
  if (pos <= ((l + r) >> 1))
    upd(l, ((l + r) >> 1), (x << 1), pos);
  else
    upd(((l + r) >> 1) + 1, r, (x << 1 | 1), pos);
  t[x] = pushup(t[(x << 1)], t[(x << 1 | 1)]);
}
node qry(int l, int r, int x, int ql, int qr) {
  if (l == ql && r == qr) return t[x];
  if (qr <= ((l + r) >> 1))
    return qry(l, ((l + r) >> 1), (x << 1), ql, qr);
  else if (ql > ((l + r) >> 1))
    return qry(((l + r) >> 1) + 1, r, (x << 1 | 1), ql, qr);
  else
    return pushup(
        qry(l, ((l + r) >> 1), (x << 1), ql, ((l + r) >> 1)),
        qry(((l + r) >> 1) + 1, r, (x << 1 | 1), ((l + r) >> 1) + 1, qr));
}
int main() {
  n = read(), q = read();
  for (register int i = 1; i <= n; i++) a[i] = read(), b[i] = read();
  build(1, n, 1);
  while (q--) {
    int opt = read();
    if (opt == 1) {
      int x = read();
      a[x] = read(), b[x] = read();
      upd(1, n, 1, x);
    } else {
      int l = read(), r = read();
      double answer = qry(1, n, 1, l, r).ans + 1;
      printf("%.10lf\n", 1.0 / answer);
    }
  }
  return 0;
}