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 int read() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + (ch ^ '0');
    ch = getchar();
  }
  return f ? x : -x;
}
const int N = 1e5 + 10;
int n, q;
double p[N], tr[N << 2][2];
inline void pushup(int p) {
  tr[p][0] = tr[p << 1][0] * tr[p << 1 | 1][0];
  tr[p][1] = tr[p << 1][1] + tr[p << 1][0] * tr[p << 1 | 1][1];
}
inline void Build(int _p, int l, int r) {
  if (l == r) {
    tr[_p][0] = tr[_p][1] = p[l];
    return;
  }
  int mid = (l + r) >> 1;
  Build(_p << 1, l, mid), Build(_p << 1 | 1, mid + 1, r);
  pushup(_p);
}
inline pair<double, double> Query(int p, int l, int r, int L, int R) {
  if (L <= l && r <= R) return make_pair(tr[p][0], tr[p][1]);
  int mid = (l + r) >> 1;
  if (L <= mid && mid < R) {
    pair<double, double> A = Query(p << 1, l, mid, L, R),
                         B = Query(p << 1 | 1, mid + 1, r, L, R), ans;
    ans.first = A.first * B.first;
    ans.second = A.second + A.first * B.second;
    return ans;
  }
  if (L <= mid) return Query(p << 1, l, mid, L, R);
  return Query(p << 1 | 1, mid + 1, r, L, R);
}
inline void Change(int p, int l, int r, int pos, double v) {
  if (l == r) {
    tr[p][0] = tr[p][1] = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    Change(p << 1, l, mid, pos, v);
  else
    Change(p << 1 | 1, mid + 1, r, pos, v);
  pushup(p);
}
int main() {
  n = read(), q = read();
  for (int i = (1), ed = (n); i <= ed; ++i)
    p[i] = 1.0 * read() / read(), p[i] = (1 - p[i]) / p[i];
  Build(1, 1, n);
  while (q--) {
    int op = read();
    if (op == 1) {
      int x = read();
      p[x] = 1.0 * read() / read(), p[x] = (1 - p[x]) / p[x];
      Change(1, 1, n, x, p[x]);
    } else {
      int l = read(), r = read();
      double res = Query(1, 1, n, l, r).second;
      printf("%.8f\n", 1.0 / (res + 1));
    }
  }
  return 0;
}