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;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &x) {
  s << "[";
  for (auto it : x) {
    s << it << ", ";
  }
  s << "]";
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, const set<T> &x) {
  s << "{";
  for (auto it : x) {
    s << it << ", ";
  }
  s << "}";
  return s;
}
template <typename U, typename V>
ostream &operator<<(ostream &s, const pair<U, V> &x) {
  s << "(" << x.first << ", " << x.second << ")";
  return s;
}
template <typename T>
bool chmax(T &x, const T &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &x, const T &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
struct node {
  long double f;
  long double h;
};
const int maxn = 1e5 + 3;
node tree[maxn * 4];
int a[maxn], b[maxn];
node combine(node fir, node sec) {
  long double g = sec.f / (1 - (1 - sec.f) * fir.h);
  node ans;
  ans.f = fir.f * g;
  ans.h = sec.h + (1 - sec.h) * fir.h * g;
  return ans;
}
void build(int v, int l, int r) {
  if (l == r) {
    tree[v].f = tree[v].h = (long double)a[l] / b[l];
    return;
  }
  int c = (l + r) / 2;
  build(v * 2 + 0, l, c);
  build(v * 2 + 1, c + 1, r);
  tree[v] = combine(tree[v * 2 + 0], tree[v * 2 + 1]);
}
void update(int v, int l, int r, int pos, long double x) {
  if (l == r) {
    tree[v].f = tree[v].h = x;
    return;
  }
  int c = (l + r) / 2;
  if (c >= pos) {
    update(v * 2 + 0, l, c, pos, x);
  } else {
    update(v * 2 + 1, c + 1, r, pos, x);
  }
  tree[v] = combine(tree[v * 2 + 0], tree[v * 2 + 1]);
}
node get_ans(int v, int l, int r, int i, int j) {
  if (i <= l && r <= j) {
    return tree[v];
  }
  int c = (l + r) / 2;
  if (c >= i && c + 1 <= j) {
    return combine(get_ans(v * 2 + 0, l, c, i, j),
                   get_ans(v * 2 + 1, c + 1, r, i, j));
  }
  if (c >= i) {
    return get_ans(v * 2 + 0, l, c, i, j);
  } else {
    return get_ans(v * 2 + 1, c + 1, r, i, j);
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
  }
  build(1, 1, n);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int pos, x, y;
      cin >> pos >> x >> y;
      update(1, 1, n, pos, (long double)x / y);
    } else {
      int l, r;
      cin >> l >> r;
      cout.precision(10);
      cout << fixed << get_ans(1, 1, n, l, r).f << "\n";
    }
  }
  return 0;
}