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 P = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
void exgcd(long long a, long long b, long long &d, long long &first,
           long long &second) {
  b                          ? exgcd(b, a % b, d, second, first),
      second -= a / b *first : first = 1, second = 0, d = a;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
const int N = 1e5 + 10;
int n, m;
struct _ {
  double L, R;
  _() {}
  _(int first, int second) { L = R = (double)first / second; }
  _(double l, double r) : L(l), R(r) {}
  _ operator+(const _ &rhs) const {
    return _(L * rhs.L / (1 - (1 - rhs.L) * R),
             rhs.R + (1 - rhs.R) * R * rhs.L / (1 - (1 - rhs.L) * R));
  }
} v[N << 2];
void build(int o, int l, int r) {
  if (l == r) {
    int first, second;
    scanf("%d%d", &first, &second);
    v[o] = _(first, second);
    return;
  }
  build((o << 1), l, ((l + r) >> 1)),
      build(((o << 1) | 1), ((l + r) >> 1) + 1, r),
      v[o] = v[(o << 1)] + v[((o << 1) | 1)];
}
void upd(int o, int l, int r, int pos, int first, int second) {
  if (l == r) return v[o] = _(first, second), void();
  if (((l + r) >> 1) >= pos)
    upd((o << 1), l, ((l + r) >> 1), pos, first, second);
  else
    upd(((o << 1) | 1), ((l + r) >> 1) + 1, r, pos, first, second);
  v[o] = v[(o << 1)] + v[((o << 1) | 1)];
}
_ qry(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return v[o];
  if (((l + r) >> 1) < ql)
    return qry(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
  if (((l + r) >> 1) >= qr) return qry((o << 1), l, ((l + r) >> 1), ql, qr);
  return qry((o << 1), l, ((l + r) >> 1), ql, qr) +
         qry(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  for (int i = 1; i <= m; ++i) {
    int op, pos, a, b;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &pos, &a, &b);
      upd(1, 1, n, pos, a, b);
    } else {
      scanf("%d%d", &a, &b);
      printf("%.12lf\n", qry(1, 1, n, a, b).L);
    }
  }
}