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;
void read(int &k) {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  k = x * f;
}
int n, m;
int a[100005], b[100005];
double p[100005];
struct segtree {
  int l, r;
  double win, lose;
} seg[100005 * 4];
void pushup(int k) {
  seg[k].win = seg[k << 1].win * seg[k << 1 | 1].win /
               (1.0 - (1.0 - seg[k << 1 | 1].win) * seg[k << 1].lose);
  seg[k].lose =
      seg[k << 1 | 1].lose +
      ((1.0 - seg[k << 1 | 1].lose) * seg[k << 1].lose * seg[k << 1 | 1].win) /
          (1.0 - (1.0 - seg[k << 1 | 1].win) * seg[k << 1].lose);
}
void build(int k, int l, int r) {
  int mid = (l + r) / 2;
  seg[k].l = l;
  seg[k].r = r;
  if (l == r) {
    seg[k].win = seg[k].lose = p[l];
    return;
  }
  build(k << 1, l, mid), build(k << 1 | 1, mid + 1, r);
  pushup(k);
}
void change(int k, int pl, double val) {
  int mid = (seg[k].l + seg[k].r) / 2;
  if (seg[k].l == pl && seg[k].r == pl) {
    seg[k].win = seg[k].lose = val;
    return;
  }
  if (pl > mid)
    change(k << 1 | 1, pl, val);
  else if (pl <= mid)
    change(k << 1, pl, val);
  pushup(k);
}
segtree query(int k, int l, int r) {
  int mid = (seg[k].l + seg[k].r) / 2;
  if (seg[k].l == l && seg[k].r == r) return seg[k];
  if (r <= mid)
    return query(k << 1, l, r);
  else if (l > mid)
    return query(k << 1 | 1, l, r);
  else {
    segtree left = query(k << 1, l, mid);
    segtree right = query(k << 1 | 1, mid + 1, r);
    segtree rnt;
    rnt.win = left.win * right.win / (1.0 - (1.0 - right.win) * left.lose);
    rnt.lose = right.lose + ((1.0 - right.lose) * left.lose * right.win) /
                                (1.0 - (1.0 - right.win) * left.lose);
    return rnt;
  }
}
int main() {
  read(n), read(m);
  for (int x, y, i = 1; i <= n; i++)
    read(x), read(y), p[i] = (double)x / (double)y;
  build(1, 1, n);
  while (m--) {
    int type;
    read(type);
    if (type == 1) {
      int pos, a, b;
      read(pos), read(a), read(b);
      change(1, pos, (double)(a) / (double)(b));
    } else {
      int a, b;
      read(a), read(b);
      printf("%.10f\n", query(1, a, b).win);
    }
  }
  return 0;
}