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>
int n, q;
long double a[100010], b[100010];
struct matrix {
  long double t[2][2];
  matrix() : t() {}
  matrix(long double a, long double b, long double c, long double d) : t() {
    t[0][0] = a, t[0][1] = b, t[1][0] = c, t[1][1] = d;
  }
  long double* operator[](int x) { return t[x]; }
  const long double* operator[](int x) const { return t[x]; }
  friend matrix operator*(matrix const& a, matrix const& b) {
    matrix c;
    c[0][0] = a[0][0] * b[0][0] + a[0][1] * b[1][0];
    c[0][1] = a[0][0] * b[0][1] + a[0][1] * b[1][1];
    c[1][0] = a[1][0] * b[0][0] + a[1][1] * b[1][0];
    c[1][1] = a[1][0] * b[0][1] + a[1][1] * b[1][1];
    return c;
  }
} tr[400010], nw;
void build(int x = 1, int l = 1, int r = n) {
  if (l == r)
    return tr[x] = matrix(0, (a[l] - b[l]) / a[l], 1, b[l] / a[l]), void();
  int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;
  build(ls, l, mid), build(rs, mid + 1, r);
  tr[x] = tr[ls] * tr[rs];
}
void update(int d, int x = 1, int l = 1, int r = n) {
  if (l == r)
    return tr[x] = matrix(0, (a[l] - b[l]) / a[l], 1, b[l] / a[l]), void();
  int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;
  if (d <= mid)
    update(d, ls, l, mid);
  else
    update(d, rs, mid + 1, r);
  tr[x] = tr[ls] * tr[rs];
}
matrix query(int pl, int pr, int x = 1, int l = 1, int r = n) {
  if (l == pl && r == pr) return tr[x];
  int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;
  if (pr <= mid) return query(pl, pr, ls, l, mid);
  if (pl > mid) return query(pl, pr, rs, mid + 1, r);
  return query(pl, mid, ls, l, mid) * query(mid + 1, pr, rs, mid + 1, r);
}
int main() {
  nw[0][0] = 0, nw[0][1] = 1;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%Lf%Lf", a + i, b + i);
  build();
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int i;
      scanf("%d", &i);
      scanf("%Lf%Lf", a + i, b + i);
      update(i);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      long double ans = 1 / (nw * query(l, r))[0][1];
      printf("%.10Lf\n", ans > 0 && ans <= 1 ? ans : (long double)0);
    }
  }
  return 0;
}