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 MAXN = 2e5 + 5;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
double getp() {
  double x, y;
  read(x), read(y);
  return x / y;
}
struct info {
  double LtoR;
  double RtoR;
};
info operator+(info a, info b) {
  info ans;
  ans.LtoR = a.LtoR * b.LtoR / (1.0 - (1.0 - b.LtoR) * a.RtoR);
  ans.RtoR = b.RtoR +
             (1.0 - b.RtoR) * a.RtoR * b.LtoR / (1.0 - (1.0 - b.LtoR) * a.RtoR);
  return ans;
}
struct SegmentTree {
  struct Node {
    int lc, rc;
    info ans;
  } a[MAXN * 2];
  int n, size, root;
  void update(int root) { a[root].ans = a[a[root].lc].ans + a[a[root].rc].ans; }
  void build(int &root, int l, int r, double *val) {
    root = ++size;
    if (l == r) {
      a[root].ans = (info){val[l], val[l]};
      return;
    }
    int mid = (l + r) / 2;
    build(a[root].lc, l, mid, val);
    build(a[root].rc, mid + 1, r, val);
    update(root);
  }
  void init(int x, double *val) {
    n = x;
    root = size = 0;
    build(root, 1, n, val);
  }
  void modify(int root, int l, int r, int pos, double val) {
    if (l == r) {
      a[root].ans = (info){val, val};
      return;
    }
    int mid = (l + r) / 2;
    if (mid >= pos)
      modify(a[root].lc, l, mid, pos, val);
    else
      modify(a[root].rc, mid + 1, r, pos, val);
    update(root);
  }
  void modify(int pos, double val) { modify(root, 1, n, pos, val); }
  info query(int root, int l, int r, int ql, int qr) {
    if (l == ql && r == qr) return a[root].ans;
    int mid = (l + r) / 2;
    if (mid >= qr)
      return query(a[root].lc, l, mid, ql, qr);
    else if (mid + 1 <= ql)
      return query(a[root].rc, mid + 1, r, ql, qr);
    else
      return query(a[root].lc, l, mid, ql, mid) +
             query(a[root].rc, mid + 1, r, mid + 1, qr);
  }
  double query(int l, int r) {
    info ans = query(root, 1, n, l, r);
    return ans.LtoR;
  }
} ST;
int n, q;
double a[MAXN];
int main() {
  read(n), read(q);
  for (int i = 1; i <= n; i++) a[i] = getp();
  ST.init(n, a);
  for (int i = 1; i <= q; i++) {
    int opt, l, r, x;
    read(opt);
    if (opt == 1) {
      read(x);
      a[x] = getp();
      ST.modify(x, a[x]);
    } else {
      read(l), read(r);
      printf("%.10lf\n", ST.query(l, r));
    }
  }
  return 0;
}