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 = 100010;
long double l[N << 2], r[N << 2];
long double p[N << 2];
void update(int node) {
  l[node] = l[node << 1] * l[node << 1 | 1] /
            (-r[node << 1] * (-l[node << 1 | 1] + 1) + 1);
  r[node] = r[node << 1 | 1] +
            (-r[node << 1 | 1] + 1) * r[node << 1] * l[node << 1 | 1] /
                (-r[node << 1] * (-l[node << 1 | 1] + 1) + 1);
}
void build(int node, int tl, int tr) {
  int m;
  if (tl == tr) {
    r[node] = l[node] = p[tl];
    return;
  }
  m = tl + tr >> 1;
  build(node << 1, tl, m);
  build(node << 1 | 1, m + 1, tr);
  update(node);
}
void modify(int node, int tl, int tr, int x, long double d) {
  int m;
  if (tl > x || tr < x) return;
  if (tl == tr) {
    l[node] = d;
    r[node] = d;
    return;
  }
  m = tl + tr >> 1;
  modify(node << 1, tl, m, x, d);
  modify(node << 1 | 1, m + 1, tr, x, d);
  update(node);
}
inline pair<long double, long double> bound(long double l1, long double r1,
                                            long double l2, long double r2) {
  pair<long double, long double> tmp;
  tmp.first = l1 * l2 / (-r1 * (-l2 + 1) + 1);
  tmp.second = r2 + (-r2 + 1) * r1 * l2 / (-r1 * (-l2 + 1) + 1);
  return tmp;
}
pair<long double, long double> query(int node, int tl, int tr, int fl, int fr) {
  int m;
  pair<long double, long double> tmp, t1, t2;
  if (fl <= tl && tr <= fr) {
    tmp = make_pair(l[node], r[node]);
    return tmp;
  }
  m = tl + tr >> 1;
  if (m + 1 <= fl) return query(node << 1 | 1, m + 1, tr, fl, fr);
  if (m >= fr) return query(node << 1, tl, m, fl, fr);
  t1 = query(node << 1, tl, m, fl, fr);
  t2 = query(node << 1 | 1, m + 1, tr, fl, fr);
  return bound(t1.first, t1.second, t2.first, t2.second);
}
int main() {
  int i, n, q;
  int x, y;
  int a, b;
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &a, &b);
    p[i] = (long double)(a) / (long double)(b);
  }
  build(1, 1, n);
  int pos, k;
  pair<long double, long double> tmp;
  long double d;
  for (i = 1; i <= q; i++) {
    scanf("%d", &pos);
    if (pos == 1) {
      scanf("%d%d%d", &k, &a, &b);
      d = (long double)(a) / (long double)(b);
      modify(1, 1, n, k, d);
    } else {
      scanf("%d%d", &x, &y);
      tmp = query(1, 1, n, x, y);
      d = tmp.first;
      printf("%.8lf\n", (double)d);
    }
  }
}